What is the answer to log_{3}(4-3^{x-1}-1)=2x-1 ?

The answer to log_{3}(4-3^{x-1}-1)=2x-1 is x=(ln((-1+sqrt(37))/2))/(ln(3))

Solve for x: log_{3}(4-3^{x-1}-1)=2x-1

The solution is x=(ln((-1+sqrt(37))/2))/(ln(3))

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

x = \frac{ln ( \frac{- 1 + 37}{2} )}{ln ( 3 )}

Decimal Notation

x = 0.84898 \dots

Solution steps

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

f (x) = g (x),

then

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

3^{l o g_{3} (4 - 3^{x - 1} - 1)} = 3^{2 x - 1}

Simplify

3^{l o g_{3} (4 - 3^{x - 1} - 1)} : 3 - 3^{x - 1}

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

Solve

3 - 3^{x - 1} = 3^{2 x - 1} : x = \frac{ln ( \frac{- 1 + 37}{2} )}{ln ( 3 )}

x = \frac{ln ( \frac{- 1 + 37}{2} )}{ln ( 3 )}

Verify Solutions:

x = \frac{ln ( \frac{- 1 + 37}{2} )}{ln ( 3 )}

True

The solution is

x = \frac{ln ( \frac{- 1 + 37}{2} )}{ln ( 3 )}

lo g_{3} (x) = 10

6 lo g_{5} (x) + 7 lo g_{2} (x) = 6

ln (x - 2) + ln (x) = 0

0.3 = lo g_{10} (x)

ln (x^{2} + 10 x + 25) = 0

What is the answer to log_{3}(4-3^{x-1}-1)=2x-1 ?
The answer to log_{3}(4-3^{x-1}-1)=2x-1 is x=(ln((-1+sqrt(37))/2))/(ln(3))
Solve for x: log_{3}(4-3^{x-1}-1)=2x-1
The solution is x=(ln((-1+sqrt(37))/2))/(ln(3))