Logarithms.

Jun 2009
186
5
Mars
Find the values of x for which:

(log x)(log x^2) + log x^3 - 5 = 0
All logs are to the base ten.

I cannpt do this.
This is what I did.
log x + log x^2 + log x^3 - 5 = 0
log x (1+x+x^2) - 5=0
log x (1+x+x^2) = 5

I got stuck there, but I am sure this method is wrong.
Please help.
Thanks
 

Debsta

MHF Helper
Oct 2009
1,284
585
Brisbane
Find the values of x for which:

(log x)(log x^2) + log x^3 - 5 = 0
All logs are to the base ten.

I cannpt do this.
This is what I did.
log x + log x^2 + log x^3 - 5 = 0
log x (1+x+x^2) - 5=0
log x (1+x+x^2) = 5

I got stuck there, but I am sure this method is wrong.
Please help.
Thanks
You need to learn your basic log rules:
1. log a + log b = log (ab) NOT log(a+b) which you tried to use in your second line. (Also in your first line your first + sign should be multiplication)
2. log a -log b = log(a/b)
and
3. log(a^n) = n log a

So: your first term:
(log x)(log x^2) = (log x)* 2log x = 2 (log x)^2

second term: (log x^3) = 3 log x

So you have:
2 (log x)^2 + 3 log x -5 = 0

(Hint: Think quadratic!!) and go from there.
 
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Oct 2009
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Find the values of x for which:

(log x)(log x^2) + log x^3 - 5 = 0
All logs are to the base ten.

I cannpt do this.
This is what I did.

log x + log x^2 + log x^3 - 5 = 0

log x (1+x+x^2) - 5=0 ----- ???How did you arrive here??

log x (1+x+x^2) = 5

I got stuck there, but I am sure this method is wrong.
Please help.
Thanks

One of basic properties of logarithms to any base is \(\displaystyle \log_ax^n=n\log_ax\) , so \(\displaystyle \log x+\log x^2+\log x^3=\log x+2\log x+3\log x=6\log x\) ...now continue and solve your problem.

Tonio
 
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Debsta

MHF Helper
Oct 2009
1,284
585
Brisbane
One of basic properties of logarithms to any base is \(\displaystyle \log_ax^n=n\log_ax\) , so \(\displaystyle \log x+\log x^2+\log x^3=\log x+2\log x+3\log x=6\log x\) ...now continue and solve your problem.

Tonio
But the sign between the first two brackets in the original equation is multiplication not addition.
 
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Reactions: Awsom Guy
Jun 2009
186
5
Mars
Tonio,
I thought I could factorise that, and I now know that this is wrong.'Thanks to Debsta.
Thanks