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

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

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

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

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

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

Decimal Notation

x = 1.23861 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

18 e^{6 x} = 2

2^{2 x} - 3 \times 2^{x} = - 2

5^{r} + 1 = 51

k, 4^{k} = n^{2}

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