What is the answer to 3*20^x=2^{x+1} ?

The answer to 3*20^x=2^{x+1} is x=(ln(2)-ln(3))/(ln(10))

Solve for x: 3*20^x=2^{x+1}

The solution is x=(ln(2)-ln(3))/(ln(10))

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3 \cdot 2 0^{x} = 2^{x + 1}

x = \frac{ln ( 2 ) - ln ( 3 )}{ln ( 10 )}

Solution steps

3 \cdot 2 0^{x} = 2^{x + 1}

Apply exponent rules

ln (3) + x ln (20) = (x + 1) ln (2)

Solve

ln (3) + x ln (20) = (x + 1) ln (2) : x = \frac{ln ( 2 ) - ln ( 3 )}{ln ( 10 )}

x = \frac{ln ( 2 ) - ln ( 3 )}{ln ( 10 )}

2^{2 x + 1} - 5 (2^{x}) - 12 = 0

(5 x - 2)^{x - 5} = (5 x - 2)^{2 x + 1}

2. 5^{- 4 r} = 6.4

x^{l o g_{10} (x)} = 1 0^{8}

9^{x} + 9 = 10 \cdot 3^{x}

What is the answer to 3*20^x=2^{x+1} ?
The answer to 3*20^x=2^{x+1} is x=(ln(2)-ln(3))/(ln(10))
Solve for x: 3*20^x=2^{x+1}
The solution is x=(ln(2)-ln(3))/(ln(10))