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    i don't understand why some log equations substitute y into the equations

    for example,

    log5(x) + 6 logx (5) = 5
    log5(x) + 6/log5(x) = 5
    let y = log5(x)
    so y + 6/y = 5....

    or

    3^(2x+1) + 5 = 16(3^x)
    let y = 3^x
    so 3y^2 + 5 = 16y ...

    why are we allowed to do that in some cases? how do i know when i should use that technique? i hope my question is clear enough, thanks in advance
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    (Original post by sadboynerd)
    i don't understand why some log equations substitute y into the equations

    for example,

    log5(x) + 6 logx (5) = 5
    log5(x) + 6/log5(x) = 5
    let y = log5(x)
    so y + 6/y = 5....

    or

    3^(2x+1) + 5 = 16(3^x)
    let y = 3^x
    so 3y^2 + 5 = 16y ...

    why are we allowed to do that in some cases? how do i know when i should use that technique? i hope my question is clear enough, thanks in advance
    It's just a technique you can use to help you solve these kinds of equations. You can use it any time you like if it's useful.

    E.g. if you had 3^x + 5^x = 3 then using y=3^x won't be much help because you'll be left with

    y + 5^x = 3


    But if you have something like 3^{2x+1} + 5 = 16(3^x) then this can be written as

    3\times (3^x)^2 + 5 = 16(3^x)

    Then if you use y=3^x you get

    3y^2 + 5 = 16y

    This is a quadratic which you should find easier to deal with than the equation above it. You just have to remember to substitute back in 3^x at the end.

    Does this make sense?
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    (Original post by notnek)
    It's just a technique you can use to help you solve these kinds of equations. You can use it any time you like if it's useful.

    E.g. if you had 3^x + 5^x = 3 then using y=3^x won't be much help because you'll be left with

    y + 5^x = 3


    But if you have something like 3^{2x+1} + 5 = 16(3^x) then this can be written as

    3\times (3^x)^2 + 5 = 16(3^x)

    Then if you use y=3^x you get

    3y^2 + 5 = 16y

    This is a quadratic which you should find easier to deal with than the equation above it. You just have to remember to substitute back in 3^x at the end.

    Does this make sense?
    it does make sense but what would be an alternative way to solving it without turning it into a quadratic?
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    (Original post by sadboynerd)
    it does make sense but what would be an alternative way to solving it without turning it into a quadratic?
    3\times (3^x)^2 + 5 = 16(3^x)

    This equation is a quadratic since you have something of the form

    3(...)^2 - 16(...) + 5 = 0

    So the only way to solve it is by using quadratic equations methods.

    But you don't have to use a substitution if you don't want to. You could just see that it is a quadratic and attempt to factorise:

    3\times (3^x)^2 -16(3^x)+5 = (3(3^x)-1)(3^x - 5)

    What I've done here is factorise just like if there was a single variable instead of 3^x e.g.

    3y^2-16y+5 = (3y-1)(y-5)

    So you can set each bracket to 0 and solve. But most people find substitution makes this process easier.
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    thank you lots for taking the time to explain!!!
 
 
 
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