How do you solve ${log}_{4} (v + 8) = {log}_{4} (- 4 v - 2)$ $log_4(v+8)=log_4(-4v-2)$ ?

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Answer 1 · 2016-09-26T10:29:44

$v = - 2$ $v=-2$

${log}_{4} (v + 8) = {log}_{4} (- 4 v - 2)$ $log_4(v+8)=log_4(-4v-2)$

$\Leftrightarrow {log}_{4} (v + 8) - {log}_{4} (- 4 v - 2) = 0$ $hArrlog_4(v+8)-log_4(-4v-2)=0$

or ${log}_{4} (\frac{v + 8}{- 4 v - 2}) = 0$ $log_4((v+8)/(-4v-2))=0$

Therefore $(\frac{v + 8}{- 4 v - 2}) = 4^{0} = 1$ $((v+8)/(-4v-2))=4^0=1$ or

$v + 8 = - 4 v - 2$ $v+8=-4v-2$ or

$v + 4 v = - 2 - 8$ $v+4v=-2-8$ or

$5 v = - 10$ $5v=-10$ or $v = - 2$ $v=-2$

How do you solve log_4(v+8)=log_4(-4v-2)log4(v+8)=log4(−4v−2)log_4(v+8)=log_4(-4v-2)?