How do you simplify $7^{{log}_{7} 9 - {log}_{7} 8}$ $7^ (log _(7)9 - log_(7)8)$ ?

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Answer 1 · 2016-05-10T11:55:34

$\frac{9}{8}$ $9/8$

Say that this equation equals $x$ $x$ , so

$7^{{log}_{7} 9 - {log}_{7} 8} = x$ $7^(log_7 9-log_7 8)=x$

Now take the ${log}_{7}$ $log_7$ of both sides to get rid of the powers,

${log}_{7} 9 - {log}_{7} 8 = {log}_{7} x$ $log_7 9-log_7 8=log_7 x$ .

We know that $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$ , so

${log}_{7} (\frac{9}{8}) = {log}_{7} x$ $log_7 (9/8)=log_7 x$

Raise both sides by the base of $7$ $7$ to remove logarithms, and find the answer

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

How do you simplify 7^ (log _(7)9 - log_(7)8)7log79−log787^ (log _(7)9 - log_(7)8)?