# Thread: a couple simple questions- logarithm and exponential

1. ## a couple simple questions- logarithm and exponential

1. lns + ln(x+2) =4
i get this down to x^2 +2x -e^4= zero. But I can't figure it out from there. Do I just use quadratic formula?

2. 3^(2x) + 3^(x+1) - 0
i get this down to x=((4-ln3)/(ln3)) * 1/3
but I can't figure out how to get the solution.

Thanks for any help in advance.

2. Originally Posted by bilbobaggins
1. lns + ln(x+2) =4
i get this down to x^2 +2x -e^4= zero. But I can't figure it out from there. Do I just use quadratic formula?

2. 3^(2x) + 3^(x+1) - 0
i get this down to x=((4-ln3)/(ln3)) * 1/3
but I can't figure out how to get the solution.

Thanks for any help in advance.
Yes quadratic formula for the first one. And I'm sure you meant lnx+ln(x+2)=4

For the second one use the substitution in the beginning 3^x=t then solve the quadratic equation

3. Originally Posted by javax
Yes quadratic formula for the first one. And I'm sure you meant lnx+ln(x+2)=4

For the second one use the substitution in the beginning 3^x=t then solve the quadratic equation
Substitute how? I can just take the x's out? Where does that leave the 2 and the +1?

4. Originally Posted by bilbobaggins
2. 3^(2x) + 3^(x+1) - 0
I'm assuming that you mean this:

$\displaystyle 3^{2x} + 3^{x + 1} = 0$

Now, $\displaystyle 3^{2x}$ is the same as $\displaystyle (3^x)^2$, and $\displaystyle 3^{x + 1}$ is the same as $\displaystyle 3(3^x)$ :

$\displaystyle (3^x)^2 + 3(3^x) = 0$

Now, let's substitute 3^x = t as javax suggested:

$\displaystyle t^2 + 3t = 0$

Hey, that looks like a quadratic equation! Solve for t:

$\displaystyle t(t + 3) = 0$

$\displaystyle t = 0$ or $\displaystyle t = -3$

But we're not done! We need to solve for x. So substitute 3^x back in for t:

$\displaystyle 3^x = 0$ or $\displaystyle 3^x = -3$

Hmm, there's no x that would satisfy either equation. The base is positive and greater than 1, so the range of the function y = 3^x would be [0, ∞). So there is no solution, unless in the original equation the RHS equals some other number instead of 0.

01

5. Originally Posted by yeongil
I'm assuming that you mean this:

$\displaystyle 3^{2x} + 3^{x + 1} = 0$

Now, $\displaystyle 3^{2x}$ is the same as $\displaystyle (3^x)^2$, and $\displaystyle 3^{x + 1}$ is the same as $\displaystyle 3(3^x)$ :

$\displaystyle (3^x)^2 + 3(3^x) = 0$

Now, let's substitute 3^x = t as javax suggested:

$\displaystyle t^2 + 3t = 0$

Hey, that looks like a quadratic equation! Solve for t:

$\displaystyle t(t + 3) = 0$

$\displaystyle t = 0$ or $\displaystyle t = -3$

But we're not done! We need to solve for x. So substitute 3^x back in for t:

$\displaystyle 3^x = 0$ or $\displaystyle 3^x = -3$

Hmm, there's no x that would satisfy either equation. The base is positive and greater than 1, so the range of the function y = 3^x would be [0, ∞). So there is no solution, unless in the original equation the RHS equals some other number instead of 0.

01
Oh, my bad. It's supposed to be a 4 and not a zero.

6. Originally Posted by yeongil
I'm assuming that you mean this:

$\displaystyle 3^{2x} + 3^{x + 1} = 0$

Now, $\displaystyle 3^{2x}$ is the same as $\displaystyle (3^x)^2$, and $\displaystyle 3^{x + 1}$ is the same as $\displaystyle 3(3^x)$ :

$\displaystyle (3^x)^2 + 3(3^x) = 4$

Now, let's substitute 3^x = t as javax suggested:

$\displaystyle t^2 + 3t = 4$

01
now
$\displaystyle t^2+3t-4=0$

$\displaystyle (t-1)(t+4)=0....t=1..or...t=-4$

but

$\displaystyle t=3^x=1...so..x=0..or...3^x=-4...that's..impossible...cuz...$

$\displaystyle take...ln..of..the..two..side$

$\displaystyle you..will...get...ln3^{x}=ln(-4)...and ...ln(-4)$

ln(-4) not defined

so you have just x=0

7. I think I figured out those 2. Thanks guys.

Here's another confusing exponential. It involves multiplication by constants.

2*49^x +11*7^x +5= 0

I'm guessing I should start by making 49 a 7^2 and moving 5 over.

2*(7^2)^x +11*7^x = -5

Do I just divide the 2 and 11 out? Or something else? Thanks again.

8. $\displaystyle 2*49^x +11*7^x +5= 0$

as you say

$\displaystyle 2(7^2)^x+11(7^x)+5=0$

$\displaystyle 2(7^x)^{2}+11(7)^x+5$

why divided the 2 and 11 ?? let t=7^x

like the question before try ...

9. Originally Posted by Amer
$\displaystyle 2*49^x +11*7^x +5= 0$

as you say

$\displaystyle 2(7^2)^x+11(7^x)+5=0$

$\displaystyle 2(7^x)^{2}+11(7)^x+5$

why divided the 2 and 11 ?? let t=7^x

like the question before try ...
Sorry, I really don't know what I'm doing when it comes to logs. Thanks for the help!

Edit: so that one has no solution then, right? I get it down to (2t+1)(t+5) which results in 2 negative numbers.

10. Originally Posted by bilbobaggins
Sorry, I really don't know what I'm doing when it comes to logs. Thanks for the help!

after putting t=7^x
we have

$\displaystyle 2t^2+11t+5=0$

$\displaystyle (2t+1)(t+5)=0$

$\displaystyle t=\frac{-1}{2}...t=-5$

$\displaystyle 7^x=\frac{-1}{2}...7^x=-5$

and in this there is no solutions for the same reason I mention before