What is the answer to solvefor m,2^{m+1}+2^m=3^{n+2}-3^n ?

The answer to solvefor m,2^{m+1}+2^m=3^{n+2}-3^n is m=(3ln(2)+ln(3)(n-1))/(ln(2))

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solvefor m,2^{m+1}+2^m=3^{n+2}-3^n

solve for

m, 2^{m + 1} + 2^{m} = 3^{n + 2} - 3^{n}

m = \frac{3 ln ( 2 ) + ln ( 3 ) ( n - 1 )}{ln ( 2 )}

Solution steps

2^{m + 1} + 2^{m} = 3^{n + 2} - 3^{n}

Apply exponent rules

m ln (2) = ln (8 \cdot 3^{n - 1})

Solve

m ln (2) = ln (8 \cdot 3^{n - 1}) : m = \frac{3 ln ( 2 ) + ln ( 3 ) ( n - 1 )}{ln ( 2 )}

m = \frac{3 ln ( 2 ) + ln ( 3 ) ( n - 1 )}{ln ( 2 )}

2^{k} + 2^{k} = 2 \cdot 2^{k}

e^{k} - 6 e^{- k} = 0

5^{- x} = 625

lo g_{2} (1 0^{x}) = 4

3^{x + 2} + 3^{x} = 810

What is the answer to solvefor m,2^{m+1}+2^m=3^{n+2}-3^n ?
The answer to solvefor m,2^{m+1}+2^m=3^{n+2}-3^n is m=(3ln(2)+ln(3)(n-1))/(ln(2))