How to solve and find the value of $log k$ equation












0












$begingroup$


I am doing a mathematics problem where the problem equation led me to below conclusion:



$$log k = 52.79$$



Now I am not sure how to solve and get value of $k$? What way we can find the value of $k$?






logarithms mathematical-physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does $log $ mean? Answer that and that answers your question. Seriously.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:31










  • $begingroup$
    $a=bimplies 10^a=10^b $.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:32










  • $begingroup$
    I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:43












  • $begingroup$
    So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:48










  • $begingroup$
    What is $10^{log k}$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:49
















0












$begingroup$


I am doing a mathematics problem where the problem equation led me to below conclusion:



$$log k = 52.79$$



Now I am not sure how to solve and get value of $k$? What way we can find the value of $k$?






logarithms mathematical-physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does $log $ mean? Answer that and that answers your question. Seriously.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:31










  • $begingroup$
    $a=bimplies 10^a=10^b $.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:32










  • $begingroup$
    I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:43












  • $begingroup$
    So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:48










  • $begingroup$
    What is $10^{log k}$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:49














0












0








0





$begingroup$


I am doing a mathematics problem where the problem equation led me to below conclusion:



$$log k = 52.79$$



Now I am not sure how to solve and get value of $k$? What way we can find the value of $k$?






logarithms mathematical-physics







share|cite|improve this question











$endgroup$




I am doing a mathematics problem where the problem equation led me to below conclusion:



$$log k = 52.79$$



Now I am not sure how to solve and get value of $k$? What way we can find the value of $k$?






logarithms mathematical-physics




logarithms mathematical-physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 7:20









Tianlalu

3,08121038




3,08121038










asked Dec 2 '18 at 3:12









ProgrammerProgrammer

2073413




2073413








  • 1




    $begingroup$
    What does $log $ mean? Answer that and that answers your question. Seriously.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:31










  • $begingroup$
    $a=bimplies 10^a=10^b $.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:32










  • $begingroup$
    I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:43












  • $begingroup$
    So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:48










  • $begingroup$
    What is $10^{log k}$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:49














  • 1




    $begingroup$
    What does $log $ mean? Answer that and that answers your question. Seriously.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:31










  • $begingroup$
    $a=bimplies 10^a=10^b $.
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:32










  • $begingroup$
    I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:43












  • $begingroup$
    So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:48










  • $begingroup$
    What is $10^{log k}$?
    $endgroup$
    – fleablood
    Dec 2 '18 at 3:49








1




1




$begingroup$
What does $log $ mean? Answer that and that answers your question. Seriously.
$endgroup$
– fleablood
Dec 2 '18 at 3:31




$begingroup$
What does $log $ mean? Answer that and that answers your question. Seriously.
$endgroup$
– fleablood
Dec 2 '18 at 3:31












$begingroup$
$a=bimplies 10^a=10^b $.
$endgroup$
– fleablood
Dec 2 '18 at 3:32




$begingroup$
$a=bimplies 10^a=10^b $.
$endgroup$
– fleablood
Dec 2 '18 at 3:32












$begingroup$
I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
$endgroup$
– Programmer
Dec 2 '18 at 3:43






$begingroup$
I am a college student and my knowledge of logs is that it gives the exponential terms that a number should be multiplied how many times - I am not sure still how to solve the question. The issue I am facing is how to solve 10 exponent 52.79
$endgroup$
– Programmer
Dec 2 '18 at 3:43














$begingroup$
So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
$endgroup$
– fleablood
Dec 2 '18 at 3:48




$begingroup$
So $log $ of $k$ is how many times you must multiply ten by so get $k$. So wouldn't that mean $k$ is what you get if you multiply $10$ by itself $52.79$?
$endgroup$
– fleablood
Dec 2 '18 at 3:48












$begingroup$
What is $10^{log k}$?
$endgroup$
– fleablood
Dec 2 '18 at 3:49




$begingroup$
What is $10^{log k}$?
$endgroup$
– fleablood
Dec 2 '18 at 3:49










1 Answer
1






active

oldest

votes


















0












$begingroup$

$log k = 52.79$



$10^{log k}=k = 10^{52.79}$



That's it you are done. But if you are asked to express $10^{52.79}$ as workable value in scientific notation then:



$10^{52.79} = 10^{52}times 10^{.79} approx 6.16595times 10^{52}$.



So that is APPROXIMATELY $61,659,500,186,148,216,632,034,834,387,861,000,000,000,000,000,000,000$. I can't do the last several values because ..... no-one cares.



That's good enough for government work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:59








  • 1




    $begingroup$
    In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 4:03






  • 1




    $begingroup$
    " is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:03










  • $begingroup$
    You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:06













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$log k = 52.79$



$10^{log k}=k = 10^{52.79}$



That's it you are done. But if you are asked to express $10^{52.79}$ as workable value in scientific notation then:



$10^{52.79} = 10^{52}times 10^{.79} approx 6.16595times 10^{52}$.



So that is APPROXIMATELY $61,659,500,186,148,216,632,034,834,387,861,000,000,000,000,000,000,000$. I can't do the last several values because ..... no-one cares.



That's good enough for government work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:59








  • 1




    $begingroup$
    In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 4:03






  • 1




    $begingroup$
    " is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:03










  • $begingroup$
    You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:06


















0












$begingroup$

$log k = 52.79$



$10^{log k}=k = 10^{52.79}$



That's it you are done. But if you are asked to express $10^{52.79}$ as workable value in scientific notation then:



$10^{52.79} = 10^{52}times 10^{.79} approx 6.16595times 10^{52}$.



So that is APPROXIMATELY $61,659,500,186,148,216,632,034,834,387,861,000,000,000,000,000,000,000$. I can't do the last several values because ..... no-one cares.



That's good enough for government work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:59








  • 1




    $begingroup$
    In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 4:03






  • 1




    $begingroup$
    " is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:03










  • $begingroup$
    You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:06
















0












0








0





$begingroup$

$log k = 52.79$



$10^{log k}=k = 10^{52.79}$



That's it you are done. But if you are asked to express $10^{52.79}$ as workable value in scientific notation then:



$10^{52.79} = 10^{52}times 10^{.79} approx 6.16595times 10^{52}$.



So that is APPROXIMATELY $61,659,500,186,148,216,632,034,834,387,861,000,000,000,000,000,000,000$. I can't do the last several values because ..... no-one cares.



That's good enough for government work.






share|cite|improve this answer











$endgroup$



$log k = 52.79$



$10^{log k}=k = 10^{52.79}$



That's it you are done. But if you are asked to express $10^{52.79}$ as workable value in scientific notation then:



$10^{52.79} = 10^{52}times 10^{.79} approx 6.16595times 10^{52}$.



So that is APPROXIMATELY $61,659,500,186,148,216,632,034,834,387,861,000,000,000,000,000,000,000$. I can't do the last several values because ..... no-one cares.



That's good enough for government work.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 4:02

























answered Dec 2 '18 at 3:54









fleabloodfleablood

68.8k22685




68.8k22685












  • $begingroup$
    Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:59








  • 1




    $begingroup$
    In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 4:03






  • 1




    $begingroup$
    " is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:03










  • $begingroup$
    You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:06




















  • $begingroup$
    Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
    $endgroup$
    – Programmer
    Dec 2 '18 at 3:59








  • 1




    $begingroup$
    In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
    $endgroup$
    – Gerry Myerson
    Dec 2 '18 at 4:03






  • 1




    $begingroup$
    " is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:03










  • $begingroup$
    You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
    $endgroup$
    – fleablood
    Dec 2 '18 at 4:06


















$begingroup$
Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
$endgroup$
– Programmer
Dec 2 '18 at 3:59






$begingroup$
Yes I understand the point but how did you calculate 10 to power .79 without calculator - is there a generic way to find it mentally or calculating?
$endgroup$
– Programmer
Dec 2 '18 at 3:59






1




1




$begingroup$
In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
$endgroup$
– Gerry Myerson
Dec 2 '18 at 4:03




$begingroup$
In the days before calculators, people published tables of logarithms (and "antilogarithms"), and you would look for $.79$ in one of those tables, and get the answer that way. Or you would take out your slide rule and use the log scale on it. Calculators have made log tables and slide rules obsolete.
$endgroup$
– Gerry Myerson
Dec 2 '18 at 4:03




1




1




$begingroup$
" is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
$endgroup$
– fleablood
Dec 2 '18 at 4:03




$begingroup$
" is there a generic way to find it easily without calculator?" Only if you have a slide rule in your closet. I didn't caluculate $10^{.79}$. I plugged it into a calculator.
$endgroup$
– fleablood
Dec 2 '18 at 4:03












$begingroup$
You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
$endgroup$
– fleablood
Dec 2 '18 at 4:06






$begingroup$
You could spend several HOURS attempting to figure out the $sqrt[100]{10^{79}}$ by approximation. BUt I don't recommend it.
$endgroup$
– fleablood
Dec 2 '18 at 4:06




















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