Math help.....help me solve these two problems for x?

Question

First off x has to be a real number, here is the first problem:

27^2x = 9^x-3

The other problem is:

log x + log (x - 3) = 1

I don't need the answers to these problems, I just really need help starting these problems. Thank You.

Answer 1

for the first problem, try expressing both sides of the equation in the same base. we are aware that 3^3=27 and 3^2=9. thus, the equation would simply turn out to be
3^3(2x) =3^2(x-3). simplifying this, we'd get:
3^6x=3^(2x-6). since they are already in the same base, all we gotta do i equate their exponents yielding:
6x=2x-6. from here, we are certain that 4x=-6, x= -3/2

for the second one, recall the property of logarithms:
log a+ log b = log ab
so, log x+ log (x-3) = log x(x-3)= log (x^2-3x) = 1. since the logarithm is in base 10, we have 10^1= x^2-3x by property: log a (base b) = c => b^c = a.

thus, we have the equation: x2-3x-10 = 0. Slving for x, we have: (x+2)(x-5)=0, x=-2,5. since x cant be negative, x=5.
there you go! good luck:D:D:D

Answer 2

For the first problem, express 27 and 9 as powers of 3
In the second problem, use this property of logs:
log a + log b = log (ab)
These are hints. Can you take it from here?