Question

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

use the exponent method to cancel out the exponents

Answer 1

I need to check what you mean by the equation. Two reasonable readings are:

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

I’ll solve both.

Case A: 3^(x−7) = 27^(2x)

Write 27 as 3^3:
3^(x−7) = (3^3)^(2x) = 3^(6x).

With the same base, equate exponents:
x − 7 = 6x
−7 = 5x
x = −7/5.

Case B: 3^x − 7 = 27^(2x)

Let y = 3^x (y > 0). Then 27^(2x) = (3^3)^(2x) = 3^(6x) = y^6. The equation becomes
y − 7 = y^6 => y^6 − y + 7 = 0.

Consider f(y) = y^6 − y + 7. f'(y) = 6y^5 − 1, which has a single real critical point y0 = (1/6)^(1/5) ≈ 0.699. Evaluating f at y0:
y0^6 = (1/6)^(6/5) = (1/6) y0 ≈ 0.1165, so
f(y0) ≈ 0.1165 − 0.699 + 7 ≈ 6.42 > 0.

Since the minimum of f on R is positive, f(y) > 0 for all real y, so there is no real solution (and hence no real x) in this case.

Summary:
- If the equation was 3^(x−7) = 27^(2x), then x = −7/5.
- If the equation was 3^x − 7 = 27^(2x), there is no real solution.

If you meant a different grouping, tell me the exact parentheses and I’ll solve that version.