How do you solve $2^{x} = 3^{2 x + 5}$ $2^x = 3^(2x+5)$ ?

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How do you solve $2^{x} = 3^{2 x + 5}$ $2^x = 3^(2x+5)$ ?

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Answer 1 · 2015-09-22T20:48:18

Refer to explanation

Explanation:

We take logarithms on both sides of the equations hence

$2^{x} = 3^{2 x + 5} \Rightarrow x log 2 = (2 x + 5) log 3 \Rightarrow x log 2 - 2 x log 3 = 5 log 3 \Rightarrow x = \frac{5 log 3}{log 2 - 2 log 3}$ $2^x=3^(2x+5)=>xlog2=(2x+5)log3=>xlog2-2xlog3=5log3=> x=(5log3)/(log2-2log3)$

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How do you solve $2^{x} = 3^{2 x + 5}$ $2^x = 3^(2x+5)$ ?

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How do you solve $2^{x} = 3^{2 x + 5}$ $2^x = 3^(2x+5)$ ?