What is the answer to solvefor x,3^{4x-5}=2^{x+3} ?

The answer to solvefor x,3^{4x-5}=2^{x+3} is x=(ln(1944))/(4ln(3)-ln(2))

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solvefor x,3^{4x-5}=2^{x+3}

solve for

x, 3^{4 x - 5} = 2^{x + 3}

x = \frac{ln ( 1944 )}{4 ln ( 3 ) - ln ( 2 )}

Decimal

x = 2.04590 \dots

Solution steps

3^{4 x - 5} = 2^{x + 3}

Apply exponent rules

(4 x - 5) ln (3) = (x + 3) ln (2)

Solve

(4 x - 5) ln (3) = (x + 3) ln (2) : x = \frac{ln ( 1944 )}{4 ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 1944 )}{4 ln ( 3 ) - ln ( 2 )}

5^{x + 3} = (\frac{1}{3})^{3 - x}

3^{1 - 2 x} = 4 x

(\frac{1}{25})^{2 x + 3} = 12 5^{9 x + 8}

5^{t} = 1637

1 6^{2 x + 1} = 4^{3 x + 3}

What is the answer to solvefor x,3^{4x-5}=2^{x+3} ?
The answer to solvefor x,3^{4x-5}=2^{x+3} is x=(ln(1944))/(4ln(3)-ln(2))