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

  • Thread starter lionely
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  • #1
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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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Answers and Replies

  • #2
ehild
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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
 
  • #3
Mentallic
Homework Helper
3,798
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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.


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?
 
  • #4
576
2


10^y = x?
 
  • #5
Mentallic
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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]
 
  • #6
576
2


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
 
  • #7
ehild
Homework Helper
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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
 
  • #8
576
2


oh.. so it's 2log x = 4 + 2log a

x^2= a^8

x=a^4?
 
  • #9
Mentallic
Homework Helper
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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]
 
  • #10
576
2


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?
 
  • #11
576
2


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?
 
  • #12
ehild
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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
 
  • #13
Mentallic
Homework Helper
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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 :smile:

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.

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]
 
  • #14
576
2


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?
 
  • #15
Mentallic
Homework Helper
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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.
 
  • #16
SammyS
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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
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 .
 
  • #17
576
2


thank you guys.
 

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