What is the answer to 2^{x+1}=3^{x-4} ?

The answer to 2^{x+1}=3^{x-4} is x=(ln(162))/(ln(3)-ln(2))

Solve for x: 2^{x+1}=3^{x-4}

The solution is x=(ln(162))/(ln(3)-ln(2))

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

x = \frac{ln ( 162 )}{ln ( 3 ) - ln ( 2 )}

Decimal Notation

x = 12.54755 \dots

Solution steps

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

Apply exponent rules

(x + 1) ln (2) = (x - 4) ln (3)

Solve

(x + 1) ln (2) = (x - 4) ln (3) : x = \frac{ln ( 162 )}{ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 162 )}{ln ( 3 ) - ln ( 2 )}

5^{x} = 0.2

lo g_{2} (2 5^{x + 3} - 1) = 2 + lo g_{2} (5^{x + 3} + 1)

\frac{2}{5} = (\frac{25}{4})^{x}

m, 144 = \frac{2 ^{m}}{3 ^{n}}

l, y^{m} - 6 y^{n} + 11 y^{l} - 6 y = 0

What is the answer to 2^{x+1}=3^{x-4} ?
The answer to 2^{x+1}=3^{x-4} is x=(ln(162))/(ln(3)-ln(2))
Solve for x: 2^{x+1}=3^{x-4}
The solution is x=(ln(162))/(ln(3)-ln(2))