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# Logarithm

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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
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#1
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Could you post a specific example???....I'm a little confused as to what you mean....

CPhill  May 8, 2017
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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