How do I Solve a Basic Logarithm Problem?

[susanto3311]

hi guys..

i need help to solve logarithm problem

how to find x?

thanks any help..

 

[soroban]

[tex]\text{Solve for }x:\;^{3x-2}\log 100 \:=\:^2\log 4.[/tex]

I've never seen logarithms written like that . . .

[tex]\begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\
& \log_{3x-2}100 \:=\:2 \\
& (3x-2)^2 \:=\:100 \\
& 3x-2 \:=\:10 \\
& 3x\:=\:12 \\
& x \:=\:4
\end{array}[/tex]

 

[topsquark]

hi guys..

i need help to solve logarithm problem

how to find x?

thanks any help..

Is this supposed to be \(\displaystyle \text{log}_{100}(3x - 2) = \text{log}_4 (2)\)?

-Dan

 

[susanto3311]

hi...

what is finally for x?

 

[susanto3311]

hi soroban...
thank, but how about this...

[tex]\begin{array}{ccc}\text{We have:} & \log_{2x-5}125 \:=\:\log_28 \\

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I've never seen logarithms written like that . . .

[tex]\begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\
& \log_{3x-2}100 \:=\:2 \\
& (3x-2)^2 \:=\:100 \\
& 3x-2 \:=\:10 \\
& 3x\:=\:12 \\
& x \:=\:4
\end{array}[/tex]

hi soroban...

 

[soroban]

[tex]\log_{2x-5}125 \:=\:\log_28 [/tex]

[tex]\begin{array}{cc}\log_{2x-5}125 \:=\: \log_28 \\
\log_{2x-5}125 \:=\:3 \\
(2x-5)^3 \:=\:125 \\
2x-5 \:=\: 5 \\
2x \:=\: 10 \\
x \:=\:5
\end{array}[/tex]
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