Can you help me solve this logarithmic equation: loga(2+x) = 2 + logax?

[Cmunro]

Hi, I've been trying to work out:

log_a(2+x) = 2 + log_ax

This is as far as I can get:

log_a(2+x) - log_ax =2
so: log_a((2+x)/x) =2
log_a((2+x)/x) =log100 (I'm not even sure where I could go with this one though).

Relevant equations include:

log x +log y = log xy
log x - log y = log x/y
log_ax=(log_cx)/log_ca

I'd really appreciate any hints you can give me. Thanks.

 

[derekjn]

It looks like you represented the 2 on the right side as a common log of 100, but the left side is log base a, so that isn't very useful in this case. Since log base a ((2+x)/x) = 2, try thinking more in terms of, "a raised to the second power is equal to ((2+x)/x)."

 

[Cmunro]

I tried to follow this trail of thought, but it didn't seem to take me anywhere:

a^2 =(2+x)/x

xa^2=2+x or alternatively a=root((2+x)/x)

..but I can't see how any of this will get rid of the a, which is ideally what I'd like to do to solve for x.

 

[Dick]

I tried to follow this trail of thought, but it didn't seem to take me anywhere:

a^2 =(2+x)/x

xa^2=2+x or alternatively a=root((2+x)/x)

..but I can't see how any of this will get rid of the a, which is ideally what I'd like to do to solve for x.

You don't want to get rid of the a. You want to solve for x in terms of a. Got that? Solve for x, not a.

 

[Cmunro]

ohhh I see! I've been quite thick really, of course I don't want to get rid of the a.

Ok so: a$$^{2}$$=2+x/x
xa$$^{2}$$-x=2
x(a$$^{2}$$-1) =2
x= 2/(a$$^{2}$$-1)

Is this right? If yes, can it be simpler?

 

[Dick]

It looks plenty simple to me.

 

[Cmunro]

Thanks for your help both of you :)