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Math Help - a couple simple questions- logarithm and exponential

  1. #1
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    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.
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  2. #2
    Member javax's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    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
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  3. #3
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    Quote Originally Posted by javax View Post
    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?
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  4. #4
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    Quote Originally Posted by bilbobaggins View Post
    2. 3^(2x) + 3^(x+1) - 0
    I'm assuming that you mean this:

    3^{2x} + 3^{x + 1} = 0

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

    (3^x)^2 + 3(3^x) = 0

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

    t^2 + 3t = 0

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

    t(t + 3) = 0

    t = 0 or t = -3

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

    3^x = 0 or 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.


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  5. #5
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    Quote Originally Posted by yeongil View Post
    I'm assuming that you mean this:

    3^{2x} + 3^{x + 1} = 0

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

    (3^x)^2 + 3(3^x) = 0

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

    t^2 + 3t = 0

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

    t(t + 3) = 0

    t = 0 or t = -3

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

    3^x = 0 or 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.
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by yeongil View Post
    I'm assuming that you mean this:

    3^{2x} + 3^{x + 1} = 0

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

    (3^x)^2 + 3(3^x) = 4

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

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


    01
    now
    t^2+3t-4=0

    (t-1)(t+4)=0....t=1..or...t=-4

    but

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

    take...ln..of..the..two..side

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

    ln(-4) not defined

    so you have just x=0
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  7. #7
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    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.
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  8. #8
    MHF Contributor Amer's Avatar
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    2*49^x +11*7^x +5= 0

    as you say

    2(7^2)^x+11(7^x)+5=0

    2(7^x)^{2}+11(7)^x+5

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

    like the question before try ...
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  9. #9
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    Quote Originally Posted by Amer View Post
    2*49^x +11*7^x +5= 0

    as you say

    2(7^2)^x+11(7^x)+5=0

    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.
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  10. #10
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    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

    2t^2+11t+5=0

    (2t+1)(t+5)=0

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

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

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