What is the answer to 7^{x+1}-2=2(7^x)+3 ?

The answer to 7^{x+1}-2=2(7^x)+3 is x=0

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7^{x+1}-2=2(7^x)+3

7^{x + 1} - 2 = 2 (7^{x}) + 3

x = 0

Solution steps

7^{x + 1} - 2 = 2 (7^{x}) + 3

Subtract

2 \cdot 7^{x}

from both sides

7^{x + 1} - 2 - 2 \cdot 7^{x} = 2 \cdot 7^{x} + 3 - 2 \cdot 7^{x}

Simplify

7^{x + 1} - 2 - 2 \cdot 7^{x} = 3

Factor

7^{x + 1} - 2 - 2 \cdot 7^{x} : 5 \cdot 7^{x} - 2

5 \cdot 7^{x} - 2 = 3

Move

2

to the right side

5 \cdot 7^{x} = 5

Divide both sides by

5

7^{x} = 1

Apply exponent rules

x ln (7) = ln (1)

Solve

x ln (7) = ln (1) : x = 0

x = 0

3^{2 x} - 6 \cdot 3^{x} + 5 = 0

2^{2 x} - 2^{x + 1} - 8 = 0

(\frac{1}{64})^{- x - 3} = 16

2^{2 x - 5} = 8^{x + 7}

2 - e^{x} = 10 e^{- x} - 5

What is the answer to 7^{x+1}-2=2(7^x)+3 ?
The answer to 7^{x+1}-2=2(7^x)+3 is x=0
Solve for x: 7^{x+1}-2=2(7^x)+3
The solution is x=0