Alan

 2log_2(x+5) = log_2(x-9) + log_2(x+53)+1

Replace 1 by log_2(2)

2log_2(x+5) = log_2(x-9) + log_2(x+53) + log_2(2)

loga + logb = logab, so

2log_2(x+5) = log_2( (x-9)*(x+53)*2)

alogb = log(b^a), so

log_2( (x+5)² ) = log_2( (x-9)*(x+53)*2)

Hence

(x+5)² = 2(x-9)(x+53)

Can you take it from here?

This should help

I'm sure you can complete the above.

As follows:

I'll leave you to finish the calculation.

Here's my solution to this:

See https://web2.0calc.com/questions/help-intermediate-algebra#r1 

Having found a and b (a = 3, b = 1) all you can do with an + b is to set it equal to 3n + 1.

The relationship  is true for all n when a = 3 and b = 1.

Add the three equations together to get

5(a + b + c) = 25  so   a + b + c = 5

Write the first equation as  2a + (a + b + c) = -3   so 2a + 5 = -3   or a = -4

Write the other two equations in a similar manner to find b and c, then multiply the three values together.

Presumably you mean find a and b (or, possibly, a + b).

I'm sure you can finish from here. 

When g(f(x)) = g(4) it means that f(x) = 4

So 4 = 2x - 3  or x = 7/2

So g(4) = 5 - 4*7/2  or g(4) = -9

f(z) = z

f(z) = -z

where z is in  R.

Both of these work.