# solving logarithmic equations

• Jun 8th 2009, 06:23 PM
extraordinarymachine
solving logarithmic equations
I need help solving for x. i am pretty new at this, and not too sure what the first couple steps would be.

log6 X + log6 (X-5)=2
• Jun 8th 2009, 06:32 PM
VonNemo19
Quote:

Originally Posted by extraordinarymachine
I need help solving for x. i am pretty new at this, and not too sure what the first couple steps would be.

log_6 X + log6 (X-5)=2

Identityloga(xy)=log_(a)x + log_(a)y

Example(a) log_(2)16 = log_(2)8 + log_(2)2

so $\log_{6}x(x-5)=2$

by the inverse property

$6^2=x(x-5)$

peace out.(Wink)
• Jun 8th 2009, 06:40 PM
I-Think
Convert the R.H.S. to logarithmic form in base 6. $(c=b^a\rightarrow a=log_{b}c)$
Use addition formula. $(log_{b}K+log_{b}L=log_b{KL})$
Drop logs.
Have fun solving, because it's all about having fun.

In action.

log6 X + log6 (X-5)=2
log6 X + log6 (X-5)=log6 36
log6 (X(X-5))=log6 36
$X^2-5X=36$

Have fun solving.
• Jun 8th 2009, 06:56 PM
mr fantastic
Quote:

Originally Posted by I-Think
Convert the R.H.S. to logarithmic form in base 6. $(c=b^a\rightarrow a=log_{b}c)$
Use addition formula. $(log_{b}K+log_{b}L=log_b{KL}$
Drop logs.
Have fun solving, because it's all about having fun.

In action.

log6 X + log6 (X-5)=2
log6 X + log6 (X-5)=log6 36
log6 (X(X-5))=log6 36
$X^2-5X=36$

Have fun solving.

@OP: Note that only solutions such that x > 5 are valid.
• Jun 8th 2009, 07:00 PM
extraordinarymachine
Thanks for all your help everyone. i think i understand now. at least to do these types of questions ha ha.