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

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

Solve for x: x-1=log_{2}(3^x)

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

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x - 1 = lo g_{2} (3^{x})

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

Solution steps

x - 1 = lo g_{2} (3^{x})

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{x - 1} = 2^{l o g_{2} (3^{x})}

Simplify

2^{l o g_{2} (3^{x})} : 3^{x}

2^{x - 1} = 3^{x}

Solve

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

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

Verify Solutions:

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

True

The solution is

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

y, lo g_{5} (y) = x

lo g_{6} (x + 1) = 2 - lo g_{6} (2)

2 lo g_{3} (4 x + 1) = 4

ln (x + 2) + \frac{x}{x + 2} = 0

lo g_{2.718} (a) = 9.06

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