How do you solve $2^{x} = 3^{x + 1}$ $2^x=3^(x+1)$ ?

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How do you solve $2^{x} = 3^{x + 1}$ $2^x=3^(x+1)$ ?

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Answer 1 · 2015-06-09T07:32:35

The value of x is "-2.71".

Explanation:

$2^{x} = 3^{x + 1}$ $2^x=3^(x+1)$
Apply logarithm with base 10 on both sides.
${log 2}^{x} = {log 3}^{x + 1}$ $log2^x=log3^(x+1)$
xlog2=(x+1)log3
By substituting the values of log2 & log3 ,we can get x value

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How do you solve $2^{x} = 3^{x + 1}$ $2^x=3^(x+1)$ ?

1 Answer

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