Help!! Can you solve this and explain the properties used? log9x+log9(x-8)=1

Question

anonymous · Answer

log(x/y) = log(x) - log(y)
log(x*y) = log(x) + log(y)
log(x^2) = 2log(x)

anonymous · Answer

first step is to distribute the 9 in log9(x-8)

anonymous · Answer

What does that result in?

anonymous · Answer

Log9x (X^2-8x)

anonymous · Answer

log9x+log9(x-8) = 1 ==> log((9x)(9x-72)) = 1

nnesha · Answer

9 is base $$\large\rm log_9~ x + \log_9 ~(x-8) =1$$

anonymous · Answer

oooh!

nnesha · Answer

use the rules which is $$\huge\rm log x + \log y = \log (x \times y )$$
like manumben gve u

nnesha · Answer

this*

anonymous · Answer

Right, and then what do I do?

anonymous · Answer

log_a(X) = b means
a^b = X

anonymous · Answer

where a is the base

nnesha · Answer

$$\huge\rm log_9  ~ x + \log_9 (x-8) $$
there is plus sign 
so you have to change this to multiplication 
remember like this 
log x + log y = log ( x times y )
theere is common base 
so log_9 ??????

anonymous · Answer

Right and I used the Product property and got Log9 (x^2-8)

anonymous · Answer

And so my equation now is Log9(x^2-8)=1 and I don't know where to go from there

nnesha · Answer

x(x-8 ) = ???

anonymous · Answer

I multiplied that through which gave me x^2-8x

nnesha · Answer

yes 8x is right |dw:1424298164973:dw|now change log to exponential form