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Thread: Is this equation right? (Logs/Algebra)

  1. #1
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    Is this equation right? (Logs/Algebra)

    So the first question is log2 (16) + log2 (8) and what I did at log2 (16) to solve it is 2x2x2x2 = 16, which means the answer is 4. And at log2 8 I did 2x2x2 = 8, which is 3. So 4+3= 7 is my answer, is this right?

    Second answer is kinda weird... The equation is log1/2 (1/8) - log1/2 (32), and I don't know how to solve this one because it's in fractions not whole numbers, can someone help me show the steps how? Btw I only have the basic kindergarten calculators. (I know 1/2 is equal to 0.5, but I'm not sure if that would work?)

    EDIT: Nevermind! I got the answer! Basically I just changed 1/2 to 0.5 and 1/8 to 0.125, I think I might solve this one!

    Third, I'm also having trouble putting this in algebra or logarithmic terms, 8^x-3 = 16^x+3
    Last edited by PeanutBrain; Nov 2nd 2017 at 12:53 PM.
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  2. #2
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    Re: Is this equation right? (Logs/Algebra)

    yes
    $\log_2(16)+\log_2(8) = 4+3=7$

    well done

    the second one is $\log_{1/2}(1/8) - \log_{1/2}(32)$ ?

    if so note that

    $\dfrac 1 8 = \left(\dfrac 1 2\right)^3$

    and

    $32 = 2^5 = \left(\dfrac 1 2\right)^{-5}$


    I assume you mean

    $8^{x-3} = 16^{x+3}$

    might as well take everything down to powers of 2

    $8^{x-3} = 2^{3(x-3)}$

    $16^{x+3} = 2^{4(x+3)}$

    and now we can say

    $3(x-3) = 4(x+3)$

    which is easily solved. (by you, or apparently by SlipEternal)
    Last edited by romsek; Nov 2nd 2017 at 01:08 PM.
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  3. #3
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    Re: Is this equation right? (Logs/Algebra)

    Quote Originally Posted by PeanutBrain View Post
    So the first question is log2 (16) + log2 (8) and what I did at log2 (16) to solve it is 2x2x2x2 = 16, which means the answer is 4. And at log2 8 I did 2x2x2 = 8, which is 3. So 4+3= 7 is my answer, is this right?

    Second answer is kinda weird... The equation is log1/2 (1/8) - log1/2 (32), and I don't know how to solve this one because it's in fractions not whole numbers, can someone help me show the steps how? Btw I only have the basic kindergarten calculators. (I know 1/2 is equal to 0.5, but I'm not sure if that would work?)

    EDIT: Nevermind! I got the answer! Basically I just changed 1/2 to 0.5 and 1/8 to 0.125, I think I might solve this one!

    Third, I'm also having trouble putting this in algebra or logarithmic terms, 8^x-3 = 16^x+3
    For the last one, you have:

    $2^{3x}-3=2^{4x}+3$
    $2^{3x}=2^{4x}+6$
    $0=2^{4x}-2^{3x}+6$
    Let $y=2^x$. Then you have
    $0=y^4-y^3+6$

    That is a 4th degree polynomial with no real solutions. So, your equation has no real solutions.

    Unless your equation is
    $8^{x-3}=16^{x+3} $

    $2^{3x-9}=2^{4x+12} $

    Implies
    $3x-9=4x+12$
    So
    $x=-21$
    Last edited by SlipEternal; Nov 2nd 2017 at 01:06 PM.
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