Question

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

use the exponent method to cancel out the exponents

show EVERY SINGLE step

Answer 1

I will assume the equation is 3^{x-7} = 27^{2x} (the minus is in the exponent). Solve by writing both sides with base 3 and then equating exponents.

1) Start: 3^{x-7} = 27^{2x}.

2) Write 27 as a power of 3: 27 = 3^3. So 27^{2x} = (3^3)^{2x}.

3) Use the power-of-a-power rule: (a^m)^n = a^{mn}. Thus (3^3)^{2x} = 3^{3·2x} = 3^{6x}.

4) Now the equation is 3^{x-7} = 3^{6x}.

5) Since the exponential function with base 3 (3 > 0, 3 ≠ 1) is one-to-one, equal powers imply equal exponents:
x - 7 = 6x.

6) Solve the linear equation: x - 7 = 6x ⇒ -7 = 5x ⇒ x = -7/5.

7) Check: Left side 3^{x-7} = 3^{-7/5 - 7} = 3^{-42/5}. Right side 27^{2x} = (3^3)^{2(-7/5)} = 3^{6(-7/5)} = 3^{-42/5}. They match.

Answer: x = -7/5.