Thread: 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.

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

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?

Originally Posted by bilbobaggins

2. 3^(2x) + 3^(x+1) - 0

I'm assuming that you mean this:



Now, is the same as , and is the same as :



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



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



or

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

or

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

Originally Posted by yeongil

I'm assuming that you mean this:



Now, is the same as , and is the same as :



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



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



or

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

or

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.

Originally Posted by yeongil

I'm assuming that you mean this:



Now, is the same as , and is the same as :



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

4" alt="t^2 + 3t = 4" />


01

now




but







ln(-4) not defined

so you have just x=0

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.

as you say





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

like the question before try ...

Originally Posted by Amer

as you say





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.

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









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