how do you solve exponential equations with added bases

Question

im confused on how you solve a question like this: $$ 3^{x+2} + 3^{x-1} = 27 $$

would you do: $$ 2(3)^{2x}-1 = 3^3 $$

but when I try this way its wrong, please help me thanks.

Answer 1

Here's a hint:

$$ 3^{x+2} = 3^3 \cdot 3^{x-1}. $$

If you use this formula to make a substitution in the equation in your question, the rest of the steps may be easier to see.

Answer 2

use that $$3^x\cdot 9+\frac{1}{3}\cdot 3^x=27$$ and we have $$3^x \cdot \frac{28}{3}=27$$

Answer 3

Let $u = 3^{x-1}$. Then $ 3^{x+2} = 3^3 \cdot u = 27 u$. Now solve $$ 27 u + u = 27 \implies u = \frac{27}{28} $$ Then $$ 3^x-1 = \frac{27}{28} \\ 3^x = \frac{55}{28} \\ x = \log_3\left( \frac{55}{28} \right) $$

Answer 4

Hint

$$3^{x+2} + 3^{x-1} = 3^x\times 3^2+3^x \times 3^{-1}=9\times 3^x+\frac 13\times 3^x=(9+\frac 13)\times 3^x=\frac{28}3\times 3^x$$ Now, define $y=3^x$, solve the equation for $y$ and use logarithms.

I am sure that you can take it from here.

Answer 5

$3^{x+2}+3^{x−1}=27$

$3^{x+2}+3^{x−1}=3^{2}\cdot3^{x}+3^{-1}\cdot3^{x}=(3^{2}+3^{-1})\cdot3^{x}$

$(3^{2}+3^{-1})\cdot3^{x}=\frac{28}{3}\cdot3^x=27$

$3^x=\frac{81}{28}$

$x=\log_3(\frac{81}{28})$