What is the answer to 7^{3x+1}=40 ?

The answer to 7^{3x+1}=40 is x=(ln(40)-ln(7))/(3ln(7))

Solve for x: 7^{3x+1}=40

The solution is x=(ln(40)-ln(7))/(3ln(7))

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7^{3 x + 1} = 40

x = \frac{ln ( 40 ) - ln ( 7 )}{3 ln ( 7 )}

Decimal Notation

x = 0.29856 \dots

Solution steps

7^{3 x + 1} = 40

Apply exponent rules

(3 x + 1) ln (7) = ln (40)

Solve

(3 x + 1) ln (7) = ln (40) : x = \frac{ln ( 40 ) - ln ( 7 )}{3 ln ( 7 )}

x = \frac{ln ( 40 ) - ln ( 7 )}{3 ln ( 7 )}

3^{2 x + 2} + 2 7^{x + 1} = 36

2^{x} + 2^{- x} = \frac{65}{8}

7^{3 x} - 5 \cdot 7^{x + 3} + 15 = 0

- 1 0^{- 9 b} = - 92.8

e^{x} \cdot e^{x} = e^{2} x

What is the answer to 7^{3x+1}=40 ?
The answer to 7^{3x+1}=40 is x=(ln(40)-ln(7))/(3ln(7))
Solve for x: 7^{3x+1}=40
The solution is x=(ln(40)-ln(7))/(3ln(7))