What is the answer to 3^{x-2}-3^{x+1}+78=0 ?

The answer to 3^{x-2}-3^{x+1}+78=0 is x=3

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3^{x-2}-3^{x+1}+78=0

3^{x - 2} - 3^{x + 1} + 78 = 0

x = 3

Solution steps

3^{x - 2} - 3^{x + 1} + 78 = 0

Factor

3^{x - 2} - 3^{x + 1} + 78 : \frac{26}{9} (- 3^{x} + 27)

\frac{26}{9} (- 3^{x} + 27) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 3^{x} + 27 = 0

Solve

- 3^{x} + 27 = 0 : x = 3

x = 3

8^{x} \cdot 4^{2 x - 1} = 32

5 e^{2 x} = 19

n, (1 - 0.01)^{n} = 0.01

4 \cdot 5^{x} - 3 = 7

8 1^{x} = \frac{1}{27}

What is the answer to 3^{x-2}-3^{x+1}+78=0 ?
The answer to 3^{x-2}-3^{x+1}+78=0 is x=3
Solve for x: 3^{x-2}-3^{x+1}+78=0
The solution is x=3