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

by lionely
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lionely
#1
Nov16-12, 01:32 AM
P: 526
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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ehild
#2
Nov16-12, 01:43 AM
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Quote Quote by lionely View Post
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
Mentallic
#3
Nov16-12, 01:43 AM
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Quote Quote by lionely View Post
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 Quote by lionely View Post
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

[tex]\log(x)=y[/tex] then what is x?

lionely
#4
Nov16-12, 01:45 AM
P: 526
(log√125 + log √27 - √8)/ log 15 - log 2 = 3/2

10^y = x?
Mentallic
#5
Nov16-12, 01:52 AM
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P: 3,540
Yes, so now do the same thing. Set the equation so that it is in the form [itex]\log(x)=y[/itex] and then make x the subject. Oh and y will be some complicated expression.

And once you've done that, remember the rules

[tex]a^{x+y}=a^xa^y[/tex]

[tex]a^{\log_{a}(x)}=x[/tex]
lionely
#6
Nov16-12, 02:01 AM
P: 526
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
ehild
#7
Nov16-12, 02:04 AM
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Quote Quote by lionely View Post


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
lionely
#8
Nov16-12, 02:13 AM
P: 526
oh.. so it's 2log x = 4 + 2log a

x^2= a^8

x=a^4?
Mentallic
#9
Nov16-12, 02:15 AM
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Quote Quote by lionely View Post
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.

[tex]\frac{\log(a)}{\log(b)}\neq \log(a)-\log(b)[/tex]

What you're thinking of is

[tex]\log\left(\frac{a}{b}\right)=\log(a)-\log(b)[/tex]
lionely
#10
Nov16-12, 02:19 AM
P: 526
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?
lionely
#11
Nov16-12, 02:43 AM
P: 526
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?
ehild
#12
Nov16-12, 04:41 AM
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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
Mentallic
#13
Nov16-12, 09:26 AM
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Quote Quote by lionely View Post
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 Quote by lionely View Post
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 [itex]\log(a)+\log(b)=\log(ab)[/itex] rule.

Quote Quote by ehild View Post
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 [itex]\sqrt{8}[/itex] should be [itex]\log(\sqrt{8})[/itex]
lionely
#14
Nov17-12, 02:19 AM
P: 526
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?
Mentallic
#15
Nov17-12, 02:47 AM
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Quote Quote by lionely View Post
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

[tex]\log(a)+\log(b)-\log(c)=\log\left(\frac{ab}{c}\right)[/tex]

for both the numerator and denominator and see if you notice any nice cancellations.
SammyS
#16
Nov17-12, 02:50 AM
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Quote Quote by ehild View Post
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 .
lionely
#17
Nov17-12, 05:04 PM
P: 526
thank you guys.


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