# Thread: Solve for x: 3^(2x) - 5(3^x) = -6

1. ## Solve for x: 3^(2x) - 5(3^x) = -6

Solve for x: 3^(2x) - 5(3^x) = -6

some explanation would be appreciated
thank you

Edit: the correct answers are 1 and 0.631, but why and how?

2. Originally Posted by incognito_301
Solve for x: 3^(2x) - 5(3^x) = -6

some explanation would be appreciated
thank you

Edit: the correct answers are 1 and 0.631, but why and how?
Here's a hint: note that $\displaystyle 3^{2x} = (3^x)^2$

now let $\displaystyle y = 3^x$

then we have $\displaystyle y^2 - 5y = -6$

that is, $\displaystyle y^2 - 5y + 6 = 0$

this is just a quadratic equation. you can solve for $\displaystyle y$ and then replace it with $\displaystyle 3^x$ and then solve for $\displaystyle x$

3. Originally Posted by incognito_301
Solve for x: 3^(2x) - 5(3^x) = -6

some explanation would be appreciated
thank you

Edit: the correct answers are 1 and 0.631, but why and how?
$\displaystyle 3^{2x} - 5(3^x) - 6 = 0$

$\displaystyle (3^{x})^2 - 5(3^x) - 6 = 0$

4. Originally Posted by incognito_301
Solve for x: 3^(2x) - 5(3^x) = -6

[snip]
I would have thought you would want someone more competent than a sap to help you.

Fortunately you got that competent help. But I wouldn't make a habit of asking for a sap to help you .....

5. LOL a sap
wow amazing forum! didnt believe you replied so fast. thanks.

alright, so

$\displaystyle A^2-5(A)-6=0$

A = -1 and A = 6

$\displaystyle 3^x = -1$
$\displaystyle 3^x = 6$

then l$\displaystyle og 6/log3 = x$
and $\displaystyle log -1/log3 = x$

i dont get the correct answers (1 and 0.631).

6. Originally Posted by incognito_301
LOL a sap
wow amazing forum! didnt believe you replied so fast. thanks.

alright, so

$\displaystyle A^2-5(A)-6=0$

A = -1 and A = 6

$\displaystyle 3^x = -1$
$\displaystyle 3^x = 6$

then l$\displaystyle og 6/log3 = x$
and $\displaystyle log -1/log3 = x$

i dont get the correct answers (1 and 0.631).
as i said, it is a +6 not -6. when you move -6 over the equal sign, it becomes positive. since you are adding 6 to both sides

but other than that, your method is correct

(by the way, log(-1) is not defined in the real numbers. you'd put "no solution" in that case)