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Math Help - exponetial eqaution

  1. #1
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    exponetial eqaution

    Hi, I have some equation that I am having trouble solving. (1) 2(2^{2x})-5(2^x)+2=0 must find x

    (2) 4^x-6(2^x)-16=0
    (3) 3^{2x+1} -26(3^x)-9=0 Must also find x for both of them. and

    I also tried this one and came up with the answer y=1 (4) \log_{2}x+\log_{x}2=2 I have to ind x
    Can anyone please help me solve these problems? I tried them but I am getting a different answer that the book has.
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  2. #2
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    Quote Originally Posted by scrible View Post
    Hi, I have some equation that I am having trouble solving. (1) 2(2^{2x})-5(2^x)+2=0 must find x

    (2) 4^x-6(2^x)-16=0
    (3) 3^{2x+1} -26(3^x)-9=0 Must also find x for both of them. and

    I also tried this one and came up with the answer y=1 (4) \log_{2}x+\log_{x}2=2 I have to ind x
    Can anyone please help me solve these problems? I tried them but I am getting a different answer that the book has.
    1. Note that this is a quadratic in 2^x. 2(2^x)^2-5(2^x)+2=0 Use the quadratic formula to find 2^x and hence x. As 2^x > 0 discard any negative solution to the quadratic.

    2. 4^x = (2^2)^x = 2^{2x}. This can then be solved in the same way as above

    3. 3^{2x+1} = 3^{2x}\cdot 3^1 = 3\cdot 3^{2x} which can also be solved in the same way as above

    4. I'm confused, what are you trying to ask?

    log_2(x)+log_x(2)=2

    Use the change of base rule to give you the same base: log_b(a) = \frac{log_c(a)}{log_c(b)}. I use base e below but any base (that isn't 1 or 0) will work

    log_2(x) = \frac{ln(x)}{ln(2)}

    log_x(2) = \frac{ln(2)}{ln(x)}

    \frac{ln(x)}{ln(2)} +  \frac{ln(2)}{ln(x)} = 2

    You can then treat that as a normal fraction question
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  3. #3
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    Quote Originally Posted by scrible View Post
    Hi, I have some equation that I am having trouble solving. (1) 2(2^{2x})-5(2^x)+2=0 must find x

    (2) 4^x-6(2^x)-16=0
    (3) 3^{2x+1} -26(3^x)-9=0 Must also find x for both of them. and

    I also tried this one and came up with the answer y=1 (4) \log_{2}x+\log_{x}2=2 I have to ind x
    Can anyone please help me solve these problems? I tried them but I am getting a different answer that the book has.
    Hi scrible,

    Here's a start. Maybe you can use this as a model for the next two.

    2(2^{2x})-5(2^x)+2=0

    (2(2^x)-1)(2^x-2)=0

    2(2^x)-1=0 \ \ or \ \ 2^x-2=0

    2(2^x)=1 \ \ or \ \ 2^x=2

    2^x=\frac{1}{2} \ \ or \ \ 2^x=2

    2^x=2^{-1} \ \ or \ \ 2^x=2^1

    x=-1 \ \ or \ \ x=1
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  4. #4
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    Quote Originally Posted by masters View Post
    Hi scrible,

    Here's a start. Maybe you can use this as a model for the next two.

    2(2^{2x})-5(2^x)+2=0

    (2(2^x)-1)(2^x-2)=0

    2(2^x)-1=0 \ \ or \ \ 2^x-2=0

    2(2^x)=1 \ \ or \ \ 2^x=2

    2^x=\frac{1}{2} \ \ or \ \ 2^x=2

    2^x=2^{-1} \ \ or \ \ 2^x=2^1

    x=-1 \ \ or \ \ x=1

    Hi it is me again I worked out this problem 3^{2x+1}-26(3^x)-9=0
    This is actual working:

    3(3^{2x}-26(3^x)-9=0

    then I said let 3^x= y

    thus 3y^2-26y-9=0

    (3y+1)(y-9)

     y= -{\frac {1}{3}} and y= 9
    Did I do every thing correctly, and is there anything more that I need ot do? The method seem correct but I am still not getting the answer...
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by scrible View Post
    Hi it is me again I worked out this problem 3^{2x+1}-26(3^x)-9=0
    This is actual working:

    3(3^{2x}-26(3^x)-9=0

    then I said let 3^x= y

    thus 3y^2-26y-9=0

    (3y+1)(y-9)

     y= -{\frac {1}{3}} and y= 9
    Did I do every thing correctly, and is there anything more that I need ot do? The method seem correct but I am still not getting the answer...
    You're correct at the moment. What you need to do next is to recall your original substitution. Also be aware that because of the nature of exponents 3^x > 0 (0 is a horizontal asymptote) which means one of your solutions is not valid
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