What is the answer to 2^{2x}=42^{x-1} ?

The answer to 2^{2x}=42^{x-1} is x=(ln(42))/(ln(21/2))

Solve for x: 2^{2x}=42^{x-1}

The solution is x=(ln(42))/(ln(21/2))

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2^{2 x} = 4 2^{x - 1}

x = \frac{ln ( 42 )}{ln ( \frac{21}{2} )}

Decimal Notation

x = 1.58956 \dots

Solution steps

2^{2 x} = 4 2^{x - 1}

Apply exponent rules

2 x ln (2) = (x - 1) ln (42)

Solve

2 x ln (2) = (x - 1) ln (42) : x = \frac{ln ( 42 )}{ln ( \frac{21}{2} )}

x = \frac{ln ( 42 )}{ln ( \frac{21}{2} )}

2^{x + 1} + 4^{x} = 80

7^{3 p - 5} = \frac{1}{( 4 9 ^{p} )}

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

5^{x - 2} \cdot 3^{2 x - 3} = 135

5^{x} = 10

What is the answer to 2^{2x}=42^{x-1} ?
The answer to 2^{2x}=42^{x-1} is x=(ln(42))/(ln(21/2))
Solve for x: 2^{2x}=42^{x-1}
The solution is x=(ln(42))/(ln(21/2))