Math Is Fun Forum

finitehelp

Member

Registered: 2009-06-21

Posts: 80

log help

Can some please help me solve these two problems step by step?

Part I:

4^x-5 = 16^x

Part II:

log 3x = log 18

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: log help

ok for your first 1:
First thing you want to do is express to same base:
4^(x-5)=16^x
4^(x-5)=4^2*x
hence now we can evaluate powers;
x-5=2x
-5=x
x=-5
Remember, when working with indices or logs, you always have to have the same base to start evaluating.

Ok for 2:
I'm assuming those logs are base 10;
log3x=log18
take log18 to other side
log(x/6)=0
x=6 as log1=0.
Hope this helps:D

Last edited by glenn101 (2009-07-01 16:00:17)

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"If your going through hell, keep going."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: log help

Hi finitehelp and glenn101;

Also,

#2 This is log(3x) = log(18). just make what is in the parentheses equal to each other, so by inspection x = 6.

Last edited by bobbym (2009-07-03 18:42:38)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: log help

interesting method there bobbym

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"If your going through hell, keep going."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: log help

Hi finitehelp and glenn101;

Also,

#2 This is log(3x) = log(18). just make what is in the parentheses equal to each other, so by inspection x = 6.

A more rigorous manner of saying this is that log is a 1-1 function, and therefore:

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

finitehelp

Member

Registered: 2009-06-21

Posts: 80

Re: log help

Thanks guys. This really helped me out. I appreciate it.