What is the answer to (0.3)^{1+x}=1.7^{2x-1} ?

The answer to (0.3)^{1+x}=1.7^{2x-1} is x=(ln(0.51))/(2ln(1.7)-ln(0.3))

Solve for x: (0.3)^{1+x}=1.7^{2x-1}

The solution is x=(ln(0.51))/(2ln(1.7)-ln(0.3))

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(0.3)^{1 + x} = 1. 7^{2 x - 1}

x = \frac{ln ( 0.51 )}{2 ln ( 1.7 ) - ln ( 0.3 )}

Decimal Notation

x = - 0.29725 \dots

Solution steps

(0.3)^{1 + x} = 1. 7^{2 x - 1}

Apply exponent rules

(1 + x) ln (0.3) = (2 x - 1) ln (1.7)

Solve

(1 + x) ln (0.3) = (2 x - 1) ln (1.7) : x = \frac{ln ( 0.51 )}{2 ln ( 1.7 ) - ln ( 0.3 )}

x = \frac{ln ( 0.51 )}{2 ln ( 1.7 ) - ln ( 0.3 )}

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

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

3 (2^{2 x + 3}) - 43 (2^{x}) + 5 = 0

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

2 e^{3 x} = 32

What is the answer to (0.3)^{1+x}=1.7^{2x-1} ?
The answer to (0.3)^{1+x}=1.7^{2x-1} is x=(ln(0.51))/(2ln(1.7)-ln(0.3))
Solve for x: (0.3)^{1+x}=1.7^{2x-1}
The solution is x=(ln(0.51))/(2ln(1.7)-ln(0.3))