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

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

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

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

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

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

Decimal Notation

x = 1.29660 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

2^{5 x + 3} = 10

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

3^{x - 1} = 5

(b^{2 x - 5})^{4} = (b^{x + 5})^{3}

e^{5 x} = 20

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