What is the answer to 4^x-4^{x-1}=3^{x+1}-3^x ?

The answer to 4^x-4^{x-1}=3^{x+1}-3^x is x=(3ln(2)-ln(3))/(2ln(2)-ln(3))

Solve for x: 4^x-4^{x-1}=3^{x+1}-3^x

The solution is x=(3ln(2)-ln(3))/(2ln(2)-ln(3))

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4^{x} - 4^{x - 1} = 3^{x + 1} - 3^{x}

x = \frac{3 ln ( 2 ) - ln ( 3 )}{2 ln ( 2 ) - ln ( 3 )}

Decimal Notation

x = 3.40942 \dots

Solution steps

4^{x} - 4^{x - 1} = 3^{x + 1} - 3^{x}

Simplify

4^{x} - 4^{x - 1} : 3 \cdot 4^{x - 1}

Simplify

3^{x + 1} - 3^{x} : 2 \cdot 3^{x}

3 \cdot 4^{x - 1} = 2 \cdot 3^{x}

Apply exponent rules

ln (3) + (x - 1) ln (4) = ln (2) + x ln (3)

Solve

ln (3) + (x - 1) ln (4) = ln (2) + x ln (3) : x = \frac{3 ln ( 2 ) - ln ( 3 )}{2 ln ( 2 ) - ln ( 3 )}

x = \frac{3 ln ( 2 ) - ln ( 3 )}{2 ln ( 2 ) - ln ( 3 )}

2^{x + 1} + 2^{x + 2} + 2^{x + 3} = 112

4^{2 x + 3} = 5^{x - 2}

x, 5 e^{x} - 9 cos x = 2

2^{2 z - 2} + 2^{z + 2} = 48

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

What is the answer to 4^x-4^{x-1}=3^{x+1}-3^x ?
The answer to 4^x-4^{x-1}=3^{x+1}-3^x is x=(3ln(2)-ln(3))/(2ln(2)-ln(3))
Solve for x: 4^x-4^{x-1}=3^{x+1}-3^x
The solution is x=(3ln(2)-ln(3))/(2ln(2)-ln(3))