How can I evaluate this logarithmic expression?
$begingroup$
I am solving a problem about recursion trees and I applied the geometric series sum formula and I have the following expression:
$$
left(frac{6}{7}right)^{log_{frac{7}{5}}(N) + 1}
$$
How can I write this in a simpler form in terms of N?
logarithms
$endgroup$
add a comment |
$begingroup$
I am solving a problem about recursion trees and I applied the geometric series sum formula and I have the following expression:
$$
left(frac{6}{7}right)^{log_{frac{7}{5}}(N) + 1}
$$
How can I write this in a simpler form in terms of N?
logarithms
$endgroup$
$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23
add a comment |
$begingroup$
I am solving a problem about recursion trees and I applied the geometric series sum formula and I have the following expression:
$$
left(frac{6}{7}right)^{log_{frac{7}{5}}(N) + 1}
$$
How can I write this in a simpler form in terms of N?
logarithms
$endgroup$
I am solving a problem about recursion trees and I applied the geometric series sum formula and I have the following expression:
$$
left(frac{6}{7}right)^{log_{frac{7}{5}}(N) + 1}
$$
How can I write this in a simpler form in terms of N?
logarithms
logarithms
edited Jan 1 at 16:15
Viktor Glombik
1,0121527
1,0121527
asked Jan 1 at 15:26
BadgerBadger
81
81
$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23
add a comment |
$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23
$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Well, if I understand right you have:
$$x:=left(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)+1}tag1$$
Using the rule:
$$text{a}^{text{b}+text{c}}=text{a}^text{b}cdottext{a}^text{c}tag2$$
We can rewrite equation $left(1right)$ as follows:
$$x=left(frac{6}{7}right)^1cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}=frac{6}{7}cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}tag3$$
Using the rule:
$$log_text{a}left(text{b}right)=frac{lnleft(text{b}right)}{lnleft(text{a}right)}tag4$$
We can rewrite equation $left(3right)$ as follows:
$$x=frac{6}{7}cdotleft(frac{6}{7}right)^frac{lnleft(text{n}right)}{lnleft(frac{7}{5}right)}tag5$$
Using the following rules:
- $$text{a}^frac{text{b}}{text{c}}=left(text{a}^frac{1}{text{c}}right)^text{b}tag6$$
- $$lnleft(frac{text{a}}{text{b}}right)=lnleft(text{a}right)-lnleft(text{b}right)tag7$$
We can rewrite equation $left(5right)$ as follows:
$$x=frac{6}{7}cdotleft(left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}right)^{lnleft(text{n}right)}tag8$$
Now, using an approximation we can write:
- $$frac{6}{7}approx0.85714tag9$$
- $$left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}approx0.63246tag{10}$$
So:
$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$
$endgroup$
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
add a comment |
$begingroup$
- Rewrite the log in base $frac{6}{7}$
- Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.
$endgroup$
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Well, if I understand right you have:
$$x:=left(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)+1}tag1$$
Using the rule:
$$text{a}^{text{b}+text{c}}=text{a}^text{b}cdottext{a}^text{c}tag2$$
We can rewrite equation $left(1right)$ as follows:
$$x=left(frac{6}{7}right)^1cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}=frac{6}{7}cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}tag3$$
Using the rule:
$$log_text{a}left(text{b}right)=frac{lnleft(text{b}right)}{lnleft(text{a}right)}tag4$$
We can rewrite equation $left(3right)$ as follows:
$$x=frac{6}{7}cdotleft(frac{6}{7}right)^frac{lnleft(text{n}right)}{lnleft(frac{7}{5}right)}tag5$$
Using the following rules:
- $$text{a}^frac{text{b}}{text{c}}=left(text{a}^frac{1}{text{c}}right)^text{b}tag6$$
- $$lnleft(frac{text{a}}{text{b}}right)=lnleft(text{a}right)-lnleft(text{b}right)tag7$$
We can rewrite equation $left(5right)$ as follows:
$$x=frac{6}{7}cdotleft(left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}right)^{lnleft(text{n}right)}tag8$$
Now, using an approximation we can write:
- $$frac{6}{7}approx0.85714tag9$$
- $$left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}approx0.63246tag{10}$$
So:
$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$
$endgroup$
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
add a comment |
$begingroup$
Well, if I understand right you have:
$$x:=left(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)+1}tag1$$
Using the rule:
$$text{a}^{text{b}+text{c}}=text{a}^text{b}cdottext{a}^text{c}tag2$$
We can rewrite equation $left(1right)$ as follows:
$$x=left(frac{6}{7}right)^1cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}=frac{6}{7}cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}tag3$$
Using the rule:
$$log_text{a}left(text{b}right)=frac{lnleft(text{b}right)}{lnleft(text{a}right)}tag4$$
We can rewrite equation $left(3right)$ as follows:
$$x=frac{6}{7}cdotleft(frac{6}{7}right)^frac{lnleft(text{n}right)}{lnleft(frac{7}{5}right)}tag5$$
Using the following rules:
- $$text{a}^frac{text{b}}{text{c}}=left(text{a}^frac{1}{text{c}}right)^text{b}tag6$$
- $$lnleft(frac{text{a}}{text{b}}right)=lnleft(text{a}right)-lnleft(text{b}right)tag7$$
We can rewrite equation $left(5right)$ as follows:
$$x=frac{6}{7}cdotleft(left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}right)^{lnleft(text{n}right)}tag8$$
Now, using an approximation we can write:
- $$frac{6}{7}approx0.85714tag9$$
- $$left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}approx0.63246tag{10}$$
So:
$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$
$endgroup$
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
add a comment |
$begingroup$
Well, if I understand right you have:
$$x:=left(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)+1}tag1$$
Using the rule:
$$text{a}^{text{b}+text{c}}=text{a}^text{b}cdottext{a}^text{c}tag2$$
We can rewrite equation $left(1right)$ as follows:
$$x=left(frac{6}{7}right)^1cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}=frac{6}{7}cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}tag3$$
Using the rule:
$$log_text{a}left(text{b}right)=frac{lnleft(text{b}right)}{lnleft(text{a}right)}tag4$$
We can rewrite equation $left(3right)$ as follows:
$$x=frac{6}{7}cdotleft(frac{6}{7}right)^frac{lnleft(text{n}right)}{lnleft(frac{7}{5}right)}tag5$$
Using the following rules:
- $$text{a}^frac{text{b}}{text{c}}=left(text{a}^frac{1}{text{c}}right)^text{b}tag6$$
- $$lnleft(frac{text{a}}{text{b}}right)=lnleft(text{a}right)-lnleft(text{b}right)tag7$$
We can rewrite equation $left(5right)$ as follows:
$$x=frac{6}{7}cdotleft(left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}right)^{lnleft(text{n}right)}tag8$$
Now, using an approximation we can write:
- $$frac{6}{7}approx0.85714tag9$$
- $$left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}approx0.63246tag{10}$$
So:
$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$
$endgroup$
Well, if I understand right you have:
$$x:=left(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)+1}tag1$$
Using the rule:
$$text{a}^{text{b}+text{c}}=text{a}^text{b}cdottext{a}^text{c}tag2$$
We can rewrite equation $left(1right)$ as follows:
$$x=left(frac{6}{7}right)^1cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}=frac{6}{7}cdotleft(frac{6}{7}right)^{log_{frac{7}{5}}left(text{n}right)}tag3$$
Using the rule:
$$log_text{a}left(text{b}right)=frac{lnleft(text{b}right)}{lnleft(text{a}right)}tag4$$
We can rewrite equation $left(3right)$ as follows:
$$x=frac{6}{7}cdotleft(frac{6}{7}right)^frac{lnleft(text{n}right)}{lnleft(frac{7}{5}right)}tag5$$
Using the following rules:
- $$text{a}^frac{text{b}}{text{c}}=left(text{a}^frac{1}{text{c}}right)^text{b}tag6$$
- $$lnleft(frac{text{a}}{text{b}}right)=lnleft(text{a}right)-lnleft(text{b}right)tag7$$
We can rewrite equation $left(5right)$ as follows:
$$x=frac{6}{7}cdotleft(left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}right)^{lnleft(text{n}right)}tag8$$
Now, using an approximation we can write:
- $$frac{6}{7}approx0.85714tag9$$
- $$left(frac{6}{7}right)^frac{1}{lnleft(7right)-lnleft(5right)}approx0.63246tag{10}$$
So:
$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$
answered Jan 1 at 15:57
JanJan
21.9k31240
21.9k31240
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
add a comment |
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
1
1
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
$begingroup$
Thank you for the detailed answer! Really helped me out with understanding this.
$endgroup$
– Badger
Jan 1 at 16:25
1
1
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
$begingroup$
@Badger You're welcome, I'm glad that I could help. And for now #HappyNewYear :)
$endgroup$
– Jan
Jan 1 at 16:32
add a comment |
$begingroup$
- Rewrite the log in base $frac{6}{7}$
- Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.
$endgroup$
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
add a comment |
$begingroup$
- Rewrite the log in base $frac{6}{7}$
- Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.
$endgroup$
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
add a comment |
$begingroup$
- Rewrite the log in base $frac{6}{7}$
- Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.
$endgroup$
- Rewrite the log in base $frac{6}{7}$
- Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.
answered Jan 1 at 15:30
user458276user458276
654211
654211
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
add a comment |
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Do you mean $$left(frac{6}{7}right)^{log_{7/5}{N}+1}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 15:36
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
$begingroup$
Yes - that’s what seemed like the correct interpretation of what the OP wrote.
$endgroup$
– user458276
Jan 1 at 15:43
add a comment |
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$begingroup$
Welcome to MSE! The expression you’ve given is ambiguous. Please check this link to learn how to format mathematical expressions.
$endgroup$
– KM101
Jan 1 at 15:30
$begingroup$
@KM101 Thank you for referring me to the link! I will make sure to use that information in my future questions.
$endgroup$
– Badger
Jan 1 at 16:23