# (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2

by lionely
Tags: &# or, 3 or 2, log&#
 P: 431 1)Without using tables, show that (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2 What i tried was (3/2 log 5 + 3/2 log 3 - 3/2 log 2)/ log 5+log 3 - log2 then from here I don't know where to take it. 2) Find the value of x if log x2/ log a^2 = log y^4/logy I tried this 2logx - 2loga = 4 log y- log y 2logx = 3logy+ 2 loga then here I get stuck..
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P: 9,264
 Quote by lionely 1)Without using tables, show that (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2 What i tried was (3/2 log 5 + 3/2 log 3 - 3/2 log 2)/ log 5+log 3 - log2 then from here I don't know where to take it. 2) Find the value of x if log x2/ log a^2 = log y^4/logy I tried this 2logx - 2loga = 4 log y- log y 2logx = 3logy+ 2 loga then here I get stuck..
You miss some parentheses.

ehild
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P: 3,323
 Quote by lionely 1)Without using tables, show that (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2 What i tried was (3/2 log 5 + 3/2 log 3 - 3/2 log 2)/ log 5+log 3 - log2 then from here I don't know where to take it.
Since you want the problem to reduce to something simple (3/2) you should combine the log terms and simplify from there.

 Quote by lionely 1)2) Find the value of x if log x2/ log a^2 = log y^4/logy I tried this 2logx - 2loga = 4 log y- log y 2logx = 3logy+ 2 loga then here I get stuck..
If we have

$$\log(x)=y$$ then what is x?

P: 431

## (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2

10^y = x?
 HW Helper P: 3,323 Yes, so now do the same thing. Set the equation so that it is in the form $\log(x)=y$ and then make x the subject. Oh and y will be some complicated expression. And once you've done that, remember the rules $$a^{x+y}=a^xa^y$$ $$a^{\log_{a}(x)}=x$$
 P: 431 I'm kind of confused if I have this log x = 3logy - 2loga I duno how to get rid of log a to make it log (x) = y
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P: 9,264
 Quote by lionely 2) Find the value of x if log x2/ log a^2 = log y^4/logy I tried this 2logx - 2loga = 4 log y- log y 2logx = 3logy+ 2 loga then here I get stuck..
log y^4/logy=4log(y)/log(y)=4

ehild
 P: 431 oh.. so it's 2log x = 4 + 2log a x^2= a^8 x=a^4?
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P: 3,323
 Quote by lionely log x2/ log a^2 = log y^4/logy I tried this 2logx - 2loga = 4 log y- log y
ehild noticed some mistakes which you should first address.

$$\frac{\log(a)}{\log(b)}\neq \log(a)-\log(b)$$

What you're thinking of is

$$\log\left(\frac{a}{b}\right)=\log(a)-\log(b)$$
 P: 431 Oh, Umm should it should be log x^2/ log a^2 = 4 2log x = 8 log a log x = 4log a 10^a^4 = 10^x x= a^4?
 P: 431 And for the first one is it (3/2log5 + 3/2log 3 - 3/2log 2)/ (log 5+ log 3) - log 2 = 1/2log log + 1/2log 3 - 1/2log2?
 HW Helper Thanks P: 9,264 I think you made some mistake when copying the problem. It should be (log√125 + log √27 - √8)/ (log 15 - log 2 )= 3/2. ehild
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P: 3,323
 Quote by lionely Oh, Umm should it should be log x^2/ log a^2 = 4 2log x = 8 log a log x = 4log a 10^a^4 = 10^x x= a^4?
Yes that's correct

 Quote by lionely And for the first one is it (3/2log5 + 3/2log 3 - 3/2log 2)/ (log 5+ log 3) - log 2 = 1/2log log + 1/2log 3 - 1/2log2?
No, again, you want to simplify that expression into a simple answer of 3/2, so what you're aiming to do is to combine each log term, not split them up further. Use the $\log(a)+\log(b)=\log(ab)$ rule.

 Quote by ehild I think you made some mistake when copying the problem. It should be (log√125 + log √27 - √8)/ (log 15 - log 2 )= 3/2. ehild
Nope the $\sqrt{8}$ should be $\log(\sqrt{8})$
 P: 431 I still don't get it combine them? But aren't they too big? shouldn't I try to get them down to like small numbers and then try to cancel out?
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P: 3,323
 Quote by lionely I still don't get it combine them? But aren't they too big? shouldn't I try to get them down to like small numbers and then try to cancel out?
Yes, combine them! If you get them down to small numbers then you'll end up with the fraction you posted earlier that cannot be cancelled further.

Use

$$\log(a)+\log(b)-\log(c)=\log\left(\frac{ab}{c}\right)$$

for both the numerator and denominator and see if you notice any nice cancellations.
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 Quote by ehild I think you made some mistake when copying the problem. It should be (log√125 + log √27 - √8)/ (log 15 - log 2 )= 3/2. ehild
Looks like a typo.

How about the first one should be:

(log√125 + log √27 - log√8)/ (log 15 - log 2 )= 3/2

As ehild said early on, you need to use sufficient parentheses .
 P: 431 thank you guys.

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