What is the answer to 3^{x+1}-8(3^{x-1})-5=0 ?

The answer to 3^{x+1}-8(3^{x-1})-5=0 is x=(ln(5))/(ln(3))+1

Solve for x: 3^{x+1}-8(3^{x-1})-5=0

The solution is x=(ln(5))/(ln(3))+1

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3^{x + 1} - 8 (3^{x - 1}) - 5 = 0

x = \frac{ln ( 5 )}{ln ( 3 )} + 1

Decimal Notation

x = 2.46497 \dots

Solution steps

3^{x + 1} - 8 (3^{x - 1}) - 5 = 0

Factor

3^{x + 1} - 8 \cdot 3^{x - 1} - 5 : 3^{x - 1} - 5

3^{x - 1} - 5 = 0

Add

5

to both sides

3^{x - 1} - 5 + 5 = 0 + 5

Simplify

3^{x - 1} = 5

Apply exponent rules

(x - 1) ln (3) = ln (5)

Solve

(x - 1) ln (3) = ln (5) : x = \frac{ln ( 5 )}{ln ( 3 )} + 1

x = \frac{ln ( 5 )}{ln ( 3 )} + 1

\frac{1}{100} = e^{- 0.15 t}

8 (2^{x - 2}) = 1 6^{x}

6^{5 x - 1} - 2 = 1

1 6^{x + 4} = 6 4^{4 x - 5}

x, 4^{x} = 8^{x + 1}

What is the answer to 3^{x+1}-8(3^{x-1})-5=0 ?
The answer to 3^{x+1}-8(3^{x-1})-5=0 is x=(ln(5))/(ln(3))+1
Solve for x: 3^{x+1}-8(3^{x-1})-5=0
The solution is x=(ln(5))/(ln(3))+1