How do you solve $log x - log (x - 8) = 1$ $log x- log(x-8)=1$ ?

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Answer 1 · 2016-01-29T10:14:16

$x = \frac{80}{9} = 8 \frac{8}{9}$ $x=80/9=8 8/9$

The given: $log x - log (x - 8) = 1$ $log x- log (x-8) = 1$

This means the common logarithm of base 10:

${log}_{10} x - {log}_{10} (x - 8) = 1$ $log_10 x-log_10(x-8) = 1$

Also ${log}_{10} 10 = 1$ $log_10 10=1$

therefore

${log}_{10} x - {log}_{10} (x - 8) = {log}_{10} 10$ $log_10 x-log_10(x-8) = log_10 10$

${log}_{10} (\frac{x}{x - 8}) = {log}_{10} 10$ $log_10 (x/(x-8))=log_10 10$

Take Antilogarithm of both sides of the equation

$A n t i log ({log}_{10} (\frac{x}{x - 8})) = A n t i log ({log}_{10} 10)$ $Antilog(log_10 (x/(x-8)))=Antilog(log_10 10)$

means

$10^{{log}_{10} (\frac{x}{x - 8})} = 10^{{log}_{10} 10}$ $10^(log_10 (x/(x-8)))=10^(log_10 10)$

means

$\frac{x}{x - 8} = 10$ $x/(x-8)=10$

solve for $x$ $x$ now

$x = 10 (x - 8)$ $x=10(x-8)$

$x = 10 x - 80$ $x=10x-80$

$x - 10 x = - 80$ $x-10x=-80$

$- 9 x = - 80$ $-9x=-80$

$\frac{- 9 x}{- 9} = \frac{- 80}{- 9}$ $(-9x)/-9=(-80)/-9$

$x = \frac{80}{9} = 8 \frac{8}{9}$ $x=80/9=8 8/9$

Have a nice day !!! from the Philippines..

How do you solve log x- log(x-8)=1logx−log(x−8)=1 log x- log(x-8)=1?