Question

`2loga_x + loga_(ax) + 3loga_(a^2x) = 0`, Find `x` ?

Answer 1

`x=10^-7" or "1/10^7`

Explanation:

The notation you have appears to be wrong, so I will assume that the first `a` in each log is the base of the log, making the equation look like this:

`2log_ax+ log_aax+3log_aa^2x=0`

Now, here are a few properties of logs that we need to look at:
1) `log_aa=1`
2) `log_ab^x=xlog_ab`
3) `log_abc=log_ab+log_ac`
4) `log_ab=log_ca/log_cblarr` this is how you change the base of a log

To our equation then:

Applying Property 3 to the second and third terms we get:
`2log_ax+ log_aa+log_ax+3(log_aa^2+log_ax)=0`

distribute that 3:
`2log_ax+log_aa+log_ax+3log_aa^2+3log_ax=0`

Now use property 2 on that 4th term:
`2log_ax+log_aa+log_ax+3*2log_aa+3log_ax=0`

Now property 1 of the 2nd and 4th terms:
`2log_ax+1+log_ax+3*2*1+3log_ax=0`

Adding all the `log_ax` terms together and all the numbers we get:
`6log_ax+7=0`
`6log_ax=-7`
`log_ax=(-7)/6`

Now we use property 4 to change the base to 10 (a log without a base is in base 10):
`log_ax=logx/loga=(-7)/6`

Looking at those fractions, we can equate the top parts and the bottom parts, and since we aren't interested in `loga` we're left with:
`logx=-7`

The last thing we need to remember is the definition of a log which is:
`log_ab=c` means that `b=a^c`

So now we have a known base for the log (10) a known quantity that the log is equal to (-7) so we just plug that in and get:
`x=10^-7" or "1/10^7`