# How do you solve logx-log2=1?

Aug 10, 2016

x = 20

#### Explanation:

Using the following $\textcolor{b l u e}{\text{laws of logarithms}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log x - \log y = \log \left(\frac{x}{y}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$
This applies to logarithms to any base.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\log}_{b} a = n \Leftrightarrow a = {b}^{n}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(B\right)$

$\text{Using (A) } \log x - \log 2 = \log \left(\frac{x}{2}\right)$

A logarithm expressed as log x , usually indicates that the base is 10.

$\text{Using (B) } {\log}_{10} \left(\frac{x}{2}\right) = 1 \Rightarrow \frac{x}{2} = {10}^{1} = 10$

Thus $\frac{x}{2} = 10 \Rightarrow x = 20$