How do you solve $- 3 = {log}_{7} x$ $-3=log_7x$ ?

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Answer 1 · 2015-11-25T18:56:47

$x = \frac{1}{343}$ $x= 1/343$

${log}_{b} p = q$ $log_b p = q$ means b^q=p#

Therefore ${log}_{7} x = - 3$ $log_7 x = -3$ means $7^{- 3} = x$ $7^(-3)= x$

$x = \frac{1}{7^{3}} = \frac{1}{343}$ $x= 1/(7^3) = 1/343$

How do you solve −3=log7x-3=log_7x?