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

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

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

The solution is x=(ln(2))/(ln(7))-1

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7^{x + 1} - 2 = 0

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

Decimal Notation

x = - 0.64379 \dots

Solution steps

7^{x + 1} - 2 = 0

Add

2

to both sides

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

Simplify

7^{x + 1} = 2

Apply exponent rules

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

Solve

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

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

1 0^{x} + 1 0^{x + 1} - 110 = 0

x, w = 3^{2 x - 5}

2433 = 3^{x}

x e^{x} = \frac{1}{2}

\frac{3 6 ^{3 + 2 n}}{6 ^{n}} = 1

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