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

The answer to log_{3}((x-1)/4)+2=-3x is x=(\W_{0}(\frac{4ln(3))/(81))+3ln(3)}{3ln(3)}

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

The solution is x=(\W_{0}(\frac{4ln(3))/(81))+3ln(3)}{3ln(3)}

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

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{81} ) + 3 ln ( 3 )}{3 ln ( 3 )}

Decimal

x = 1.01563 \dots

Solution steps

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

Move

2

to the right side

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

f (x) = g (x),

then

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

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

Simplify

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

\frac{x - 1}{4} = 3^{- 3 x - 2}

Multiply both sides by

4

\frac{x - 1}{4} \cdot 4 = 3^{- 3 x - 2} \cdot 4

Simplify

x - 1 = 3^{- 3 x - 2} \cdot 4

Solve

x - 1 = 3^{- 3 x - 2} \cdot 4 : x = \frac{W _{0} ( \frac{4 l n ( 3 )}{81} ) + 3 ln ( 3 )}{3 ln ( 3 )}

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{81} ) + 3 ln ( 3 )}{3 ln ( 3 )}

Verify Solutions:

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{81} ) + 3 ln ( 3 )}{3 ln ( 3 )}

True

The solution is

x = \frac{W _{0} ( \frac{4 l n ( 3 )}{81} ) + 3 ln ( 3 )}{3 ln ( 3 )}

ln (x) + ln (x^{2}) = 6

- 30 = 20 lo g_{10} (10 \cdot m)

y, 3 ln (x) = ln (y)

20 lo g_{10} (\frac{x}{2}) = a

lo g_{10} (x) = \frac{3}{4}

What is the answer to log_{3}((x-1)/4)+2=-3x ?
The answer to log_{3}((x-1)/4)+2=-3x is x=(\W_{0}(\frac{4ln(3))/(81))+3ln(3)}{3ln(3)}
Solve for x: log_{3}((x-1)/4)+2=-3x
The solution is x=(\W_{0}(\frac{4ln(3))/(81))+3ln(3)}{3ln(3)}