What is the answer to 4^{x+1}-5^{x+1}=(0.8)(4^x) ?

The answer to 4^{x+1}-5^{x+1}=(0.8)(4^x) is x=(ln(5)-ln(3.2))/(2ln(2)-ln(5))

Solve for x: 4^{x+1}-5^{x+1}=(0.8)(4^x)

The solution is x=(ln(5)-ln(3.2))/(2ln(2)-ln(5))

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4^{x + 1} - 5^{x + 1} = (0.8) (4^{x})

x = \frac{ln ( 5 ) - ln ( 3.2 )}{2 ln ( 2 ) - ln ( 5 )}

Decimal Notation

x = - 2

Solution steps

4^{x + 1} - 5^{x + 1} = (0.8) (4^{x})

Move

0.8 \cdot 4^{x}

to the left side

3.2 \cdot 4^{x} - 5^{x + 1} = 0

Move

5^{x + 1}

to the right side

3.2 \cdot 4^{x} = 5^{x + 1}

Apply exponent rules

ln (3.2) + x ln (4) = (x + 1) ln (5)

Solve

ln (3.2) + x ln (4) = (x + 1) ln (5) : x = \frac{ln ( 5 ) - ln ( 3.2 )}{2 ln ( 2 ) - ln ( 5 )}

x = \frac{ln ( 5 ) - ln ( 3.2 )}{2 ln ( 2 ) - ln ( 5 )}

1.73 = 3^{x}

5^{- 3 x - 1} = 25

x^{l o g_{10} (x)} = 100

3^{x} = \frac{1}{729}

a^{2 x - 5} = a^{15}

What is the answer to 4^{x+1}-5^{x+1}=(0.8)(4^x) ?
The answer to 4^{x+1}-5^{x+1}=(0.8)(4^x) is x=(ln(5)-ln(3.2))/(2ln(2)-ln(5))
Solve for x: 4^{x+1}-5^{x+1}=(0.8)(4^x)
The solution is x=(ln(5)-ln(3.2))/(2ln(2)-ln(5))