Question

Find the exact solution to the equation. 10-log^8 (x+2) =9

x = 12

x = -6

x = 10

x = 6

Answer 1

`\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^(log_a x)=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 10-log_8(x+2)=9\implies 10=9+log_8(x+2)\implies 1=log_8(x+2) \\\\\\ 8^1=8^(log_8(x+2))\implies 8=x+2\implies 6=x`