Question #99695

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Question #99695

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Answer 1 · 2017-06-18T21:58:17

The logs more or less disappear!

Explanation:

${(x^{2} + 6)}^{{log}_{2} (x)} = {(5 x)}^{{log}_{2} (x)}$ $(x^2 + 6)^(log_2(x)) = (5x)^(log_2(x))$

A special problem. If it were possible for the bases to be either positive or negative, we might encounter a case involving an absolute value. For now, don't worry about what that is. Why? Because the base "5x" is negative if and only if x is negative.

We know that x cannot be negative because ${log}_{2} (x)$ $log_2(x)$ is undefined unless $x > 0$ $x > 0$ . Also, $x^{2} + 6$ $x^2 + 6$ is always positive; therefore $5 x$ $5x$ must be positive.

In this case...
${(x^{2} + 6)}^{{log}_{2} (x)} = {(5 x)}^{{log}_{2} (x)}$ $(x^2 + 6)^(log_2(x)) = (5x)^(log_2(x))$
if and only if the bases are equal. That is,
$x^{2} + 6 = 5 x$ $x^2 + 6 = 5x$
Solve this as
$x^{2} - 5 x + 6 = 0$ $x^2 -5x + 6 = 0$
by using factoring.

Finish it off now. There are two solutions.

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Question #99695

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