What is the answer to 7^{x+1}=2^{3x-2} ?

The answer to 7^{x+1}=2^{3x-2} is x=(ln(28))/(3ln(2)-ln(7))

Solve for x: 7^{x+1}=2^{3x-2}

The solution is x=(ln(28))/(3ln(2)-ln(7))

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

x = \frac{ln ( 28 )}{3 ln ( 2 ) - ln ( 7 )}

Decimal Notation

x = 24.95446 \dots

Solution steps

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

Apply exponent rules

(x + 1) ln (7) = (3 x - 2) ln (2)

Solve

(x + 1) ln (7) = (3 x - 2) ln (2) : x = \frac{ln ( 28 )}{3 ln ( 2 ) - ln ( 7 )}

x = \frac{ln ( 28 )}{3 ln ( 2 ) - ln ( 7 )}

5^{2 x + 1} - 6 (5^{x}) + 1 = 0

2^{5 x + 4} = 6

4 = (\frac{1}{2})^{x}

1 5^{2 x + 3} = 1 5^{x - 7}

3 e^{x} = 1 - 2 e^{0.5 x}

What is the answer to 7^{x+1}=2^{3x-2} ?
The answer to 7^{x+1}=2^{3x-2} is x=(ln(28))/(3ln(2)-ln(7))
Solve for x: 7^{x+1}=2^{3x-2}
The solution is x=(ln(28))/(3ln(2)-ln(7))