Condensing a logarithmic expression

[khavar]

I made it through my laws of logarithms studies this morning, except for one problem.

Condense: log_ax + log_a20 - log(x²+1) - log_a10

log_a(20x/10) - log(x²+1)
My next step is a change of base for: log(x²+1)
log_a2x - (log_a(x²+1)/log_a10)
I'm not sure what to do next from here.
Perhaps:
log_a2x/((x²+1)/10) = log_a(2x(10) / (x²+1))

The answer in my book is: log_a(2x/(x²+1))

Thank you for the help. My next section is "The Number e and Natural Logarithms". Excited for the progress!

 

[Deleted member 4993]

I made it through my laws of logarithms studies this morning, except for one problem.

Condense: log_ax + log_a20 - log_a(x²+1) - log_a10 .............. → to match the answer in the back, the problem must be this.

log_a(20x/10) - log(x²+1)
My next step is a change of base for: log(x²+1)
log_a2x - (log_a(x²+1)/log_a10)
I'm not sure what to do next from here.
Perhaps:
log_a2x/((x²+1)/10) = log_a(2x(10) / (x²+1))

The answer in my book is: log_a(2x/(x²+1))

Thank you for the help. My next section is "The Number e and Natural Logarithms". Excited for the progress!

.