+0  
 
0
160
2
avatar

If I have a logarithm and I'm solving for X, and my log base is a variable, do I have to solve for the log base first before I can solve for X?

Guest May 8, 2017
Sort: 

2+0 Answers

 #1
avatar+76929 
0

 

Could you post a specific example???....I'm a little confused as to what you mean....

 

 

 

cool cool cool

CPhill  May 8, 2017
 #2
avatar+6810 
0

Example:

\(\log_x(5) = \dfrac{1}{4}\\ x^{\log_x(5)}=x^{1/4}\quad\text{Unless x = 0, if you have x = 0, you did something wrong.}\\ x^{1/4}=5\\ x = 5^4 = 625 \)

OR

Gonna use the same question

\(\log_x(5)=\dfrac{1}{4}\\ \dfrac{1}{\log_5x}=\dfrac{1}{4} \quad\leftarrow \boxed{\log_ab=\dfrac{1}{\log_ba}}\\ \log_5x = 4\\ x = 5^4 = 625\)

MaxWong  May 8, 2017

23 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details