How do you solve $9 {log 2}^{x} = 18$ $9 log 2^x =18$ ?

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Answer 1 · 2016-03-07T08:36:39

${log 2}^{x} = 2 \to x \cdot log 2 = 2 \to x = \frac{2}{log 2}$ $log2^x =2->x*log2 = 2->x=2/log2$

First, divide the 9 over to the other side then bring the exponent x to the front. Finally, solve for x by dividing log 2 to the other side.

How do you solve 9log2x=189 log 2^x =18?