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

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

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

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

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

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

Decimal

x = - 1.55329 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

e^{x} = e^{x}

2 \cdot 3^{5 x} = 1

8 1^{x^{2}} = 2 7^{x}

4^{x} = 47

4^{x} = 27

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