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

The answer to 2^{x+1}=3^{x-2} is x=(ln(18))/(ln(3)-ln(2))

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

The solution is x=(ln(18))/(ln(3)-ln(2))

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

x = \frac{ln ( 18 )}{ln ( 3 ) - ln ( 2 )}

Decimal Notation

x = 7.12853 \dots

Solution steps

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

Apply exponent rules

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

Solve

(x + 1) ln (2) = (x - 2) ln (3) : x = \frac{ln ( 18 )}{ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 18 )}{ln ( 3 ) - ln ( 2 )}

e^{9 x} = 5

8^{- 3 x} = \frac{1}{4}

1000000 = 1 0^{1.5 d}

10 = e^{x}

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

What is the answer to 2^{x+1}=3^{x-2} ?
The answer to 2^{x+1}=3^{x-2} is x=(ln(18))/(ln(3)-ln(2))
Solve for x: 2^{x+1}=3^{x-2}
The solution is x=(ln(18))/(ln(3)-ln(2))