Question: Solve {127^(x+2) - 6*3^(3x+3)} / {3^x*9^(x+2) }=7
The denominator is:
3^x*9^(x+2) = 3^x*3^[2*(x+2)] = 3^x*3^(2x+4) = 3^(3x+4)
In the numerator,
6*3^(3x+3) = 2*3*3^(3x+3) = 2*3^(3x+4)
The equation now is: {127^(x+2) - 2*3^(3x+4)} / {3^(3x+4) } = 7
cross-multiplying gives us,
127^(x+2) - 2*3^(3x+4) = 7*3^(3x+4)
127^(x+2) = 7*3^(3x+4) + 2*3^(3x+4)
127^(x+2) = 9*3^(3x+4)
127^(x+2) = 3^(3x+6)
Taking logs,
(x+2).ln(127) = (3x+6).ln(3)
But ln(127) and ln(3) are just constants, C1 and C2, say. Then
(x+2)*C1 = (x+2)*C2', where C2' = 3*C2
and this equation is satisfied for x = -2.
Answer: x = -2
You could also work it out longhand.
x.ln(127) + 2.ln(127) = 3x.ln(3) + 6.ln(3)
x{ln(127) - ln(3^3)} = ln(3^6) - ln(127^2)
x.ln(127/27) = ln(729/16,129)
x = ln(729/16,129) / ln(127/27) = -3.0967/1.54835 = -2
i.e. x = -2