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    hi y'all
    i'm revising for a test on ib (sl) algebra and i'm, like, moments from ripping all of my hair out
    also i missed quite a bit of lessons on this so this question might be real stupid but bear with me

    q1: solve for t: 4^(t)=1/1024
    i went with multiplying both sides by 1024, then (t)log4096=1 and t=1/log4096. i got 0.277, the answer is apparently -5.
    q2: solve for x: 2^(2x)-2^(x+1)+8 = 0
    i let 2^(x)=y and tried to solve it like a quadratic but it wouldn't factorize.
    the answers are probably really obvious so sorry in advance for posting this!!
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    For Q1:
    4^t can be written as 2^2t
    1/1024 cab be written as 2^(-10)
    Therefore you get 2t=-10
    t=-5
    For Q2:
    XXXXXXX
    Hope it helps

    Just noticed my mistake sorry. So I am deleting the answer for q2.

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    (Original post by Rorsy)
    hi y'all
    i'm revising for a test on ib (sl) algebra and i'm, like, moments from ripping all of my hair out
    also i missed quite a bit of lessons on this so this question might be real stupid but bear with me

    q1: solve for t: 4^(t)=1/1024
    i went with multiplying both sides by 1024, then (t)log4096=1 and t=1/log4096. i got 0.277, the answer is apparently -5.
    q2: solve for x: 2^(2x)-2^(x+1)+8 = 0
    i let 2^(x)=y and tried to solve it like a quadratic but it wouldn't factorize.
    the answers are probably really obvious so sorry in advance for posting this!!


    1) BEDMAS you can't multiply something with an exponent that isn't the same base i.e. 4^{t} = \frac{1}{1024} \neq 4096^{t} = 1

    just take logs from the start

    2) if it doesn't factorise, use the quadratic formula
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    (Original post by Rorsy)
    hi y'all
    i'm revising for a test on ib (sl) algebra and i'm, like, moments from ripping all of my hair out
    also i missed quite a bit of lessons on this so this question might be real stupid but bear with me

    q1: solve for t: 4^(t)=1/1024
    i went with multiplying both sides by 1024, then (t)log4096=1 and t=1/log4096. i got 0.277, the answer is apparently -5.
    q2: solve for x: 2^(2x)-2^(x+1)+8 = 0
    i let 2^(x)=y and tried to solve it like a quadratic but it wouldn't factorize.
    the answers are probably really obvious so sorry in advance for posting this!!
    also OP are you sure you've posted q2 correctly?

    i have good reason to believe the solution does not exist
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    (Original post by Rorsy)
    hi y'all
    i'm revising for a test on ib (sl) algebra and i'm, like, moments from ripping all of my hair out
    also i missed quite a bit of lessons on this so this question might be real stupid but bear with me

    q1: solve for t: 4^(t)=1/1024
    i went with multiplying both sides by 1024, then (t)log4096=1 and t=1/log4096. i got 0.277, the answer is apparently -5.
    q2: solve for x: 2^(2x)-2^(x+1)+8 = 0
    i let 2^(x)=y and tried to solve it like a quadratic but it wouldn't factorize.
    the answers are probably really obvious so sorry in advance for posting this!!
    For Q2)
    It is a quadratic equation for 2^x
    2^(2x)=(2^x)^2
    2^(x+1)=2*2^x
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    (Original post by ztibor)
    For Q2)
    It is a quadratic equation for 2^x
    2^(2x)=(2^x)^2
    2^(x+1)=2*2^x
    aaaaaaand real solutions do not exist
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    (Original post by boner in jeans)
    aaaaaaand real solutions do not exist
    No,it exists (substitute 2^x=y -> y^2-2y-8=0 will be the equation)
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    (Original post by ztibor)
    No,it exists
    x=2
    ?

    f(x) = 2^{2x} - 2^{x+1} + 8

    f(2) = 16

    yes, and look at that quadratic y^{2} - 2y + 8 = 0
    b^{2} - 4ac < 0

    no real solutions
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    (Original post by boner in jeans)
    ?

    f(x) = 2^{2x} - 2^{x+1} + 8

    f(2) = 16
    Ok. The constant +8 and I wrote -8)
 
 
 
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