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    Given that 4x=82-x, find the value of x. Without using logs!

    Forgot how to do this simple question -_- (I know you have to write both numbers in the form of 2)
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    4^x=8^(2-x)
    2^(2x)=2^(3(2-x))
    2^(2x)=2^(6-3x)
    2x = 6 - 3x
    x=1.2
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    (Original post by The Clockwork Apple)
    4^x=8^(2-x)
    2^(2x)=2^(3(2-x))
    2^(2x)=2^(6-3x)
    2x = 6 - 3x
    x=1.2
    I actually just got it right after I posted this, but thank you anyway!
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    (Original post by rm_27)
    I actually just got it right after I posted this, but thank you anyway!
    That's OK
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    (Original post by The Clockwork Apple)
    2^(2x)=2^(6-3x)
    2x = 6 - 3x
    oOHoh but u used logs there
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    (Original post by CancerousProblem)
    oOHoh but u used logs there
    No
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    (Original post by The Clockwork Apple)
    No
    Yes u did that's how u got from the line there to the line on the bottom u just didnt write it
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    (Original post by CancerousProblem)
    Yes u did that's how u got from the line there to the line on the bottom u just didnt write it
    a^b = a^c => b = c, no? That's just a property of exponentiation, is it not?
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    (Original post by StrangeBanana)
    a^b = a^c => b = c, no?
    yeah if u take logs then it is

    f(a) = f(b) does not imply a = b, only the other way around.

    you have to take logs
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    (Original post by CancerousProblem)
    yeah if u take logs then it is
    :| I don't think you need logs to prove that result
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    (Original post by CancerousProblem)
    f(a) = f(b) does not imply a = b, only the other way around.

    you have to take logs
    It does if f is a bijective function like 2^x
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    (Original post by StrangeBanana)
    Not in general, it doesn't, but in this case it does
    because the result holds for logarithims
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    (Original post by CancerousProblem)
    because the result holds for logarithims
    see 2 posts above, there is no need to use logarithms
 
 
 
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