How do you solve $10^(-12x)+6=100$ ?

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Answer 1 · 2016-07-16T18:41:38

$x = -0.1644$

$10^(-12x)+6=100$

$10^(-12x)=94$

As the variable is in the index, we will use logs.

$log10^(-12x)=log94$

$-12x xxlog10=log94 " "log 10 = 1$

$x =log94/(-12)$

$x = -0.1644$

How do you solve 10^(-12x)+6=10010^(-12x)+6=100?