Logarithms and solving for x

[Louise Johnson]

Hello I am working through problems solving for x in my Logarithms section. This first problem I can solve (I hope its correct as it works?) however the second below it has got me stumped. Can anyone help me out?
Thank you
sincerly
Louise

Question: Solve for x

\(\displaystyle \L\\3(3^{x - 1} ) = \frac{1}{{27}}\)

my answer:
\(\displaystyle \log 3 + (x - 1)\log 3 = \log \frac{1}{{27}}\)

\(\displaystyle (x - 1)\log 3 = \frac{{\log \frac{1}{{27}} - \log 3}}{{\log 3}}\)

\(\displaystyle x = \frac{{\log \frac{1}{{27}} - \log 3}}{{\log 3}} + 1 = 3\)

Question #2

\(\displaystyle \L\\ln e^{\sqrt {x - 2} } = 6\)

 

[mark07]

First one should be -3, since

\(\displaystyle \L \log \frac{1}{27} = \log 3^{-3} = -3 \log 3\)

#2)

\(\displaystyle \L \ln e^{\sqrt {x - 2} }
= ( \sqrt {x - 2} ) \ln e
= \sqrt {x - 2}\)

Solve

\(\displaystyle \L \sqrt {x - 2} = 6\)

 

[Denis]

\(\displaystyle \L\\\3(3^{x - 1} ) = \frac{1}{{27}}\)

Easier this way, Louise:
3^(x - 1) = (1 / 27) / 3
3^(x - 1) = 1 / 81
x - 1 = log(1 / 81) / log(3)
x - 1 = -4
x = -3

 

[Louise Johnson]

Thanks Mark and Dennis I feel pretty comfortable using both methods although the second problem was a tough one I've finally got that sorted.
Thank you again
Sincerly
Louise

 

[soroban]

Hello, Louise!

I read your problem differently.
And it can be solved without logs . . .


To me, it looks like: \(\displaystyle \L\:3^{(3^{x - 1})} \;= \;\frac{1}{{27}}\)

So we have: \(\displaystyle \L\:\left(3^3)^{x-1} \:=\:27^{-1}\)

. . . . . . . . . . .\(\displaystyle \L27^{x-1}\:=\:27^{-1}\)

The bases are equal, so the exponents are equal:

. . \(\displaystyle \L x\,-\,1\:=\:-1\;\;\Rightarrow\;\;\fbox{x\:=\:0}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If it really was: \(\displaystyle \:3\cdot3^{x-1}\;=\;\frac{1}{27} \:=\:\frac{1}{3^{-3}}\)

. . we have: \(\displaystyle \:3^x \;=\;3^{-3}\)

Therefore: \(\displaystyle \:x\,=\,-3\)

 

[pka]

If the problem were \(\displaystyle 3^{\left( {3^{x - 1} } \right)} = \frac{1}{{27}}\) then there would be no solution. Because \(\displaystyle 3^{\left( {3^{x - 1} } \right)} = \frac{1}{{27}}\quad \Rightarrow \quad 3^{x - 1} = - 3\) which is clearly impossible.