How do you solve $4 log x - log x = 64$ $4 log x - log x = 64$ ?

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Answer 1 · 2015-12-21T20:29:09

$x = 10^{\frac{64}{3}}$ $x=10^(64/3)$

$4 log x - log x = 64$ $4logx-logx=64$

$3 log x = 64$ $3logx=64$

$log x = \frac{64}{3}$ $logx=64/3$

Recall that $log x = {log}_{10} x$ $logx=log_10x$ .

Exponentiate both sides to undo the logarithm.

$10^{{log}_{10} x} = 10^{\frac{64}{3}}$ $10^(log_10x)=10^(64/3)$

$x = 10^{\frac{64}{3}}$ $x=10^(64/3)$

$x \approx 2.1544 \times 10^{21}$ $x~~2.1544xx10^21$

How do you solve 4logx−logx=644 log x - log x = 64?