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I need help solving $−\log_2(x−1) \; \log_2(3x−4)>0$.

The logarithms have a base number of 2.. I have tried many times but I can't find the answer.

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    Examine each of $\log_{2}(x-1)$ and $\log_{2}(3x-4)$ separately: for which $x$ is the logarithm positive? Negative? – 2017-01-24

3 Answers 3

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Logs are positive when the argument is greater than $1$ and the argument must be positive. You need one of $x-1$ and $3x-4$ to be between $0$ and $1$ and the other to be greater than $1$.

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You want the product of the two logarithms to be negative. This is true exactly when one of them is positive and the other is negative. Now split into two cases according to which of them is the positive one.

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If we want $-\log_2(x-1)\log_2(3x-4)$ to be positive, we need $\log_2(x-1)$ OR $\log_2(3x-4)$ to be negative, but not both. Therefore, we can determine the regions where they are positive and negative.

$\log_2(3x-4)<0 \iff 2^{(\log_2(3x-4))} < 2^0 \iff 3x-4 < 1 \iff x < \frac{5}{3}$

and similarly

$\log_2(x-1)<0 \iff 2^{(\log_2(x-1))} < 2^0 \iff x-1 < 1 \iff x < 2\iff x < \frac{6}{3}$.

Therefore, we need $x$ to satisfy exactly one of these inequalities above. Now, if $x<5/3$ it satisfies both of these inequalities, and if $x\geq 6/3$ it satisfies neither of these inequalities. Therefore, we need $5/3 < x < 6/3$, noting that if $x = 5/3$ or $x = 6/3$ one of the $\log$s will be $0$ and the product will be $0$.