Solving an exponential function with a sum

Question

I have to solve equations of this kind to $x$:

$3^{2x-1} + 1 = 28 \cdot 3^{x-2}$

I don't get the trick to eliminate the $+1$ in the equation. Can someone show me how I can solve this? Thanks!

Answer 1

If we set $$y = 3^{x-2} \implies y^2 = 3^{2x-4} = \frac{1}{27}3^{2x-1},$$ We can see that

$$3^{2x-1} + 1 = 28 \cdot 3^{x-2}$$ $$\implies 27y^2 + 1 = 28y$$ $$\implies 27y^2 - 28y + 1 = 0$$ $$\implies (27y-1)(y - 1) = 0$$ $$\implies y = \frac{1}{27} \vee y = 1$$ $$\implies 3^{x-2} = \frac{1}{27} \vee 3^{x-2} = 1$$ $$\implies x = -1 \vee x = 2.$$

Answer 2

You could use substitution.

Let $u=3^{x-2}$, so $u^2=3^{2x-4}$ and $3^{3}u^2=3^{2x-1}$

So now we just need to solve $3^{3}u^2+1=28u$.

And the that would give us $u=\frac{1}{27}$ or $u=1$

$3^{x-2}=\frac{1}{27}\Rightarrow x=-1$

$3^{x-2}=1 \Rightarrow x=2$

Answer 3

Hint:

write the equation as: $$ \frac{3^{2x}}{3}+1=28\frac{3^x}{9} $$

than use the substitution $y=3^x$