# How do you solve log_10 7=log_10 x-log_10 2?

Oct 10, 2015

$x = 14$

#### Explanation:

For two logarithms of the same base, ${\log}_{a} M = {\log}_{a} N \Rightarrow M = N$

Using the Quotient Law, ${\log}_{a} x - {\log}_{a} y = {\log}_{a} \left(\frac{x}{y}\right)$

So,

${\log}_{10} 7 = {\log}_{10} \left(\frac{x}{2}\right)$
$7 = \frac{x}{2}$
$x = 7 \cdot 2$
$x = 14$

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ABR May I suggest another approach:

Logs are indices in another form as such adding indices is another representation of multiplication. Like wise, subtraction is another representation of division. Consequently:

It is given that $\log 7 = \log x - \log 2$

( it does not matter a bout what the base is in this context)

$A n t i \log \left(\log 7\right) = 7 , A n t i \log \left(\log x\right) = x , A n t i \log \left(\log 2\right) = 2$

Thus $\log 7 = \log x - \log 2 \text{ is equivalent to } 7 = \frac{x}{2}$

From this it is a simple matter of manipulation to find $x$