logarithm simplification

[5ugxr]

I'm having trouble with simplifying this logarithmic expression to sqrt6, anything I try keeps getting messed up because of the different bases and the different exponents. I also tried rewriting 24 as 3 * 2^3 and 54 as 2 * 3^3 but I wasn't able to simplify

 

[mario99]

I think that you have a typo and the expression should be:

[imath]\displaystyle \frac{3^{\log(54)/\log(9/4)}}{2^{\log(54)/\log(9/4)}} = 3\sqrt{6}[/imath]

 

[Dr.Peterson]

View attachment 38369
I'm having trouble with simplifying this logarithmic expression to sqrt6, anything I try keeps getting messed up because of the different bases and the different exponents. I also tried rewriting 24 as 3 * 2^3 and 54 as 2 * 3^3 but I wasn't able to simplify

Wolfram Alpha does give 2.4494897..., which is [imath]\sqrt{6}[/imath]; but it doesn't simplify it to that expression, suggesting that this isn't particularly easy ...

 

[5ugxr]

I think that you have a typo and the expression should be:

[imath]\displaystyle \frac{3^{\log(54)/\log(9/4)}}{2^{\log(54)/\log(9/4)}} = 3\sqrt{6}[/imath]

 

[5ugxr]

Wolfram Alpha does give 2.4494897..., which is [imath]\sqrt{6}[/imath]; but it doesn't simplify it to that expression, suggesting that this isn't particularly easy ...

Yea, I was trying to find some online calculators to simplify it but they all keep giving me answers in logarithmic form...
My teacher said I have to use exponent and logarithm laws to simplify (that's a given...) but I'm not seeing any way to do it

 

[khansaheb]

View attachment 38369
I'm having trouble with simplifying this logarithmic expression to sqrt6, anything I try keeps getting messed up because of the different bases and the different exponents. I also tried rewriting 24 as 3 * 2^3 and 54 as 2 * 3^3 but I wasn't able to simplify

What do you get - in log form & NOT in decimal form - when you simplify:

\(\displaystyle \dfrac{\log{24}}{\log({\frac{9}{4})}}\)

 

[5ugxr]

What do you get - in log form & NOT in decimal form - when you simplify:

\(\displaystyle \dfrac{\log{24}}{\log({\frac{9}{4})}}\)

Log base 9/4 of 24?

 

[khansaheb]

Log base 9/4 of 24?

Incorrect - how did you get that?

Start with:

log (9/4) ............ the base of "log" here is assumed to be 10 , according to common convention

log(9/4) = log(9) - log(4) = 2 * log(3) - 2 * log(2).................. and

log(24) = log(3*2 * 2*2) = log(3) + 3*log(2)

 

[5ugxr]

Incorrect - how did you get that?

Start with:

log (9/4) ............ the base of "log" here is assumed to be 10 , according to common convention

log(9/4) = log(9) - log(4) = 2 * log(3) - 2 * log(2).................. and

log(24) = log(3*2 * 2*2) = log(3) + 3*log(2)

Oh, yes I understand that
I thought this was expanding it haha, I can see your steps tho
I was a bit confused, it was log(a)/log(b) so I was just thinking change of base rule

 

[BigBeachBanana]

This is unorthodox of the site but here's my solution.

Spoiler

[math]\frac{3^{\log(24)/ \log(9/4)}}{2^{\log(54)/\log(9/4)}} = x; \text{where } x > 0 [/math][math]\dfrac{3 \log(2) + \log(3)}{2 \log(3/2)}\log 3 -\dfrac{\log(2) + 3 \log(3)}{2 \log(3/2)}\log 2= \log x[/math][math]\dfrac{(\log(3))^2-(\log(2))^2}{2 \log(3/2)}= \log x[/math][math]\dfrac{(\log(3)+\log(2))(\log(3)-\log(2))}{2(\log(3)-\log(2))}= \log x[/math][math]\dfrac{\log(3)+\log(2)}{2}= \log x[/math][math]\log(6) = \log(x^2)[/math][math]6=x^2[/math][math]x=\sqrt{6}[/math]

 

[mario99]

This is unorthodox of the site but here's my solution.

Spoiler

[math]\frac{3^{\log(24)/ \log(9/4)}}{2^{\log(54)/\log(9/4)}} = x; \text{where } x > 0 [/math][math]\dfrac{3 \log(2) + \log(3)}{2 \log(3/2)}\log 3 -\dfrac{\log(2) + 3 \log(3)}{2 \log(3/2)}\log 2= \log x[/math][math]\dfrac{(\log(3))^2-(\log(2))^2}{2 \log(3/2)}= \log x[/math][math]\dfrac{(\log(3)+\log(2))(\log(3)-\log(2))}{2(\log(3)-\log(2))}= \log x[/math][math]\dfrac{\log(3)+\log(2)}{2}= \log x[/math][math]\log(6) = \log(x^2)[/math][math]6=x^2[/math][math]x=\sqrt{6}[/math]

I came up to the same idea, but you were faster.

 

[5ugxr]

I was confused for a minute trying to see how that was diff of squares but I can see it now by making a substitution. Would've been stuck on this question for a while
Thanks guys!!

 

[khansaheb]

This is unorthodox of the site but here's my solution.

Spoiler

[math]\frac{3^{\log(24)/ \log(9/4)}}{2^{\log(54)/\log(9/4)}} = x; \text{where } x > 0 [/math][math]\dfrac{3 \log(2) + \log(3)}{2 \log(3/2)}\log 3 -\dfrac{\log(2) + 3 \log(3)}{2 \log(3/2)}\log 2= \log x[/math][math]\dfrac{(\log(3))^2-(\log(2))^2}{2 \log(3/2)}= \log x[/math][math]\dfrac{(\log(3)+\log(2))(\log(3)-\log(2))}{2(\log(3)-\log(2))}= \log x[/math][math]\dfrac{\log(3)+\log(2)}{2}= \log x[/math][math]\log(6) = \log(x^2)[/math][math]6=x^2[/math][math]x=\sqrt{6}[/math]

Much simpler than my way. I was simplifying the given expression and replace log(3)= A and log(2)=B and then simplify further as algebraic expression. After 3 pages - came to the same point.

 

[BigBeachBanana]

After 3 pages - came to the same point.

Though my way is shorter, 3 pages is the way to go.

 

[mario99]

Much simpler than my way. I was simplifying the given expression and replace log(3)= A and log(2)=B and then simplify further as algebraic expression. After 3 pages - came to the same point.

Professor Khan,

Can you please show us your three pages of solution?

 

[khansaheb]

Professor Khan,

Can you please show us your three pages of solution?

BBB has already shown the correct (and pithy) solution and I do not want to waste time.