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

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

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

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

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

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

Decimal Notation

x = - 5.19089 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

6 4^{3 x} = 8^{24}

2^{6 - 3 x} = 8^{x + 1}

7^{x - 3} \times 98 = 14

2^{2 x} - 12 (2^{x}) + 35 = 0

1 0^{- 3 x} \cdot 1 0^{x} = \frac{1}{10}

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