How can I evaluate this logarithmic expression?












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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:



$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$






share|cite|improve this answer









$endgroup$









  • 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



















1












$begingroup$


  1. Rewrite the log in base $frac{6}{7}$

  2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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:



$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$






share|cite|improve this answer









$endgroup$









  • 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
















1












$begingroup$

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:



$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$






share|cite|improve this answer









$endgroup$









  • 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














1












1








1





$begingroup$

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:



$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$






share|cite|improve this answer









$endgroup$



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:



$$xapprox0.85714cdot0.63246^{lnleft(text{n}right)}tag{11}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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











1












$begingroup$


  1. Rewrite the log in base $frac{6}{7}$

  2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$


  1. Rewrite the log in base $frac{6}{7}$

  2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$


  1. Rewrite the log in base $frac{6}{7}$

  2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.






share|cite|improve this answer









$endgroup$




  1. Rewrite the log in base $frac{6}{7}$

  2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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