What is the answer to 7^{x-1}-2^x=0 ?

The answer to 7^{x-1}-2^x=0 is x=(ln(7))/(ln(7)-ln(2))

Solve for x: 7^{x-1}-2^x=0

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

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7^{x - 1} - 2^{x} = 0

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

Decimal Notation

x = 1.55329 \dots

Solution steps

7^{x - 1} - 2^{x} = 0

Add

2^{x}

to both sides

7^{x - 1} - 2^{x} + 2^{x} = 0 + 2^{x}

Simplify

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 )}

9^{x} - 4^{x} = 6^{x}

7 e^{10 x} = 12

4^{x} - 3 \times 2^{x + 1} + 8 = 0

3^{2 x - 1} + 3^{2 x} = 108

1 6^{x - 1} = 64

What is the answer to 7^{x-1}-2^x=0 ?
The answer to 7^{x-1}-2^x=0 is x=(ln(7))/(ln(7)-ln(2))
Solve for x: 7^{x-1}-2^x=0
The solution is x=(ln(7))/(ln(7)-ln(2))