How do you solve $8^(-2-x)=431$ ?

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Answer 1 · 2017-02-05T14:50:57

$x=-4.9172$

As from definition of log $a^m=b$ implies $log_a b=m$

Also $log_a b=logb/loga$ where $log$ without mentioning base means logarithm to the base $10$ and we can get this from tables.

$8^(-2-x)=431$ means $log_8 431=-2-x$

or $-2-x=log431/log8=2.6345/0.9031=2.9172$

Hence, $x=-2-2.9172=-4.9172$

How do you solve 8^(-2-x)=4318^(-2-x)=431?