Question

Solve the following equation. 3^x−7=27^2x

but use the exponent thing where you cancel out the exponents
for example you should use x-7=2x

Answer 1

Could you confirm how the equation is meant to be parsed? There are two different reasonable readings:

A) 3^(x−7) = 27^(2x) (i.e. the exponent on 3 is x−7), or
B) 3^x − 7 = 27^(2x) (i.e. 7 is subtracted from 3^x).

If you meant A (which fits your "cancel out the exponents" comment), solve like this:

1. Write both sides with base 3: 27 = 3^3, so
3^(x−7) = (3^3)^(2x) = 3^(6x).

2. Now the bases are equal and positive, so equate exponents:
x − 7 = 6x.

3. Solve for x:
−7 = 5x => x = −7/5.

Check: 3^(−7/5 − 7) = 3^(−42/5) and 27^(2x) = (3^3)^(−14/5) = 3^(−42/5), so it matches.

If you meant B (3^x − 7 = 27^(2x)), there is no real solution. Rewriting gives 3^x − 3^(6x) = 7; the left-hand side has a maximum value < 1 for real x, so it never reaches 7.

Tell me which parsing you intended if you want more detail.