MHB 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]

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

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

\begin{array}{ccc}\text{We have:} &amp; \log_{3x-2}100 \:=\:\log_24 \\<br /> &amp; \log_{3x-2}100 \:=\:2 \\<br /> &amp; (3x-2)^2 \:=\:100 \\<br /> &amp; 3x-2 \:=\:10 \\<br /> &amp; 3x\:=\:12 \\<br /> &amp; x \:=\:4<br /> \end{array}

 

[topsquark]

hi guys..

i need help to solve logarithm problem

how to find x?

thanks any help..

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

-Dan

 

[susanto3311]

hi...

what is finally for x?

 

[susanto3311]

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

\begin{array}{ccc}\text{We have:} &amp; \log_{2x-5}125 \:=\:\log_28 \\<br /> <br /> - - - Updated - - -<br /> <br /> <blockquote data-attributes="member: 703424" data-quote="soroban" data-source="post: 6750174" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> soroban said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I&#039;ve never seen logarithms written like that . . .<br /> <br /> \begin{array}{ccc}\text{We have:} &amp;amp; \log_{3x-2}100 \:=\:\log_24 \\&lt;br /&gt; &amp;amp; \log_{3x-2}100 \:=\:2 \\&lt;br /&gt; &amp;amp; (3x-2)^2 \:=\:100 \\&lt;br /&gt; &amp;amp; 3x-2 \:=\:10 \\&lt;br /&gt; &amp;amp; 3x\:=\:12 \\&lt;br /&gt; &amp;amp; x \:=\:4&lt;br /&gt; \end{array} </div> </div> </blockquote><br /> hi soroban...

 

[soroban]

\log_{2x-5}125 \:=\:\log_28

\begin{array}{cc}\log_{2x-5}125 \:=\: \log_28 \\<br /> \log_{2x-5}125 \:=\:3 \\<br /> (2x-5)^3 \:=\:125 \\<br /> 2x-5 \:=\: 5 \\<br /> 2x \:=\: 10 \\<br /> x \:=\:5<br /> \end{array}
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