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Can you solve this without trial and error? Watch

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    5^x +4^x =8
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    x must be less than 1
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    (Original post by the bear)
    x must be less than 1
    I know, but can I figure the solution without trial and error?
    The exact value
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    x is bigger than 0
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    (Original post by mariejuana)
    I know, but can I figure the solution without trial and error?
    The exact value
    Change the bases.
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    Hay ho
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    (Original post by RDKGames)
    Change the bases.
    I did and it gave me a weird answer.
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    Rty
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    (Original post by RDKGames)
    What was the weird answer?
    i changed 4^x t0 2^2x. changed 8 to 2^3
    used logs to change 5 t0 a power of 2

    2^x= 5
    x=log5/log2
    so 2^log5/log2 x + 2^2x= 2^3
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    (Original post by mariejuana)
    i changed 4^x t0 2^2x. changed 8 to 2^3
    used logs to change 5 t0 a power of 2

    2^x= 5
    x=log5/log2
    so 2^log5/log2 x + 2^2x= 2^3
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be \displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.
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    (Original post by RDKGames)
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.

    Thank youuu!!
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    (Original post by RDKGames)
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be \displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.


    What about 5^x + 2*5^1-x =7 ?
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    (Original post by mariejuana)
    What about 5^x + 2*5^1-x =7 ?
    You can easily solve that and find an exact solution.
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    (Original post by RDKGames)
    You can easily solve that and find an exact solution.
    how ? xD
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    (Original post by mariejuana)
    how ? xD
    Multiply through by 5^x to get (5^x)^2+10=7(5^x) which is a quadratic.
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    (Original post by RDKGames)
    Multiply through by 5^x to get (5^x)^2+10=7(5^x) which is a quadratic.

    i don't understand it at allllllllllllll
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    (Original post by mariejuana)
    i don't understand it at allllllllllllll
    Ok, it's literally nothing more difficult than solving a quadratic. Let y=5^x then you have y^2+10=7y which is the same as y^2-7y+10=0 which I'm sure you can solve before convering y back to 5^x
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    (Original post by mariejuana)
    i don't understand it at allllllllllllll
    i do. doesn#t matter i was beingstupud
 
 
 
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