How do you solve ${log}_{7} (2 x^{2} - 1) = {log}_{7} 4 x$ $log_7(2x^2 - 1) = log_7 4x$ ?

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Answer 1 · 2015-11-24T16:06:42

$2 + \sqrt{6}$ $2+sqrt6$

Note that if ${log}_{a} b = {log}_{a} c,$ $log_ab=log_ac,$ then $b = c$ $b=c$ .

Because of this, we can say that $2 x^{2} - 1 = 4 x$ $2x^2-1=4x$ .

$2 x^{2} - 4 x - 1 = 0$ $2x^2-4x-1=0$

$x = \frac{4 \pm \sqrt{16 + 8}}{2}$ $x=(4+-sqrt(16+8))/2$

$x = 2 \pm \sqrt{6}$ $x=2+-sqrt6$

However, the only answer is the positive one, $2 + \sqrt{6}$ $2+sqrt6$ , sincein ${log}_{a} b$ $log_ab$ , $b > 0$ $b>0$ .

How do you solve log_7(2x^2 - 1) = log_7 4xlog7(2x2−1)=log74xlog_7(2x^2 - 1) = log_7 4x?