math

Q1:solve log(5)x-log(25)(x+10) = 0.5

Q2:If 2log(a)x=1+log(a) (7x-10a) ,find x in terms of a.

Q3:Find x for which 27x3^lgx = 9^1+lg(x-20)


Q4:Find x in terms of a and c ,given that log(√ a)(1/x)+log(a)x +log(a^2) x +log (a^4)x=c

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asked by Anonymous
  1. since 25=5^2, log(25) = 1/2 log(5) and we have

    log(5)x - 1/2 log(5)(x+1) = 0.5
    or,
    2log(5)x - log(5)(x+10) = 1
    log(5)(x^2 / (x+10)) = 1
    x^2/(x+10) = 5
    x^2 = 5x+50
    x^2-5x-50 = 0
    (x-10)(x+5) = 0
    x=10, since log(-5) is not defined.
    ----------------------------------
    2log(a)x=1+log(a) (7x-10a)
    since log(a)a = 1,
    log(a) x^2 = log(a) (a(7x-10a))
    x^2 = a(7x-10a)
    x^2 = 7ax - 10a^2
    x^2 - 7ax + 10a^2 = 0
    (x-5a)(x-2a) = 0
    x = 2a or 5a
    -------------------------------
    27x3^lgx = 9^1+lg(x-20)
    Not sure what this means, but if you meant

    27(3^lgx) = 9^(1+lg(x-20))
    3^3 * 3^lgx = 3^(2 + 2lg(x-20))
    1+lgx = 2+2lg(x-20)
    lg10 + lgx = lg100 + lg(x-20)^2
    10x = 100(x-20)^2
    10x = 100x^2 - 4000x + 400
    100x^2 - 4010x + 400 = 0
    10(10x-1)(x-40) = 0
    x = 1/10 or 40
    -------------------------------------
    log(√a)(1/x)+log(a)x +log(a^2)x +log(a^4)x=c
    2log(a)(1/x) + log(a)x + 1/2 log(a)x + 4log(a)x = c
    (1/x^2)(x)(√x)(x^4) = a^c
    x^(7/2) = a^c
    x = a^(2/7 c)

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    posted by Steve
  2. For the Qn3 ,how u get 3^1+lgx from 3^3 *3^ lgx
    ? why not 3^ 3+lgx

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    posted by Anonymous
  3. And the ans for Qn4 should be x=a^(-4c)...

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    posted by Anonymous
  4. btw,teacher,how to do this Question? log(5)(5-4x)=log(√5)(2-x) ?

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    posted by Anonymous
  5. Q4c:
    1/x^2*x*sqrt(x)*sqrt(sqrt(x))=a^c;
    1/x^(1/4) = a^c
    x=a^(-4c)

    For:
    log5(5-4x)=log√5(2-x)

    log5(5-4x)=log5(2-x)²
    5-4x=(2-x)²
    x=±1

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    posted by MathMate
  6. Way to watch, MathMate. I missed the 1/x^2 and read it as x^2. Simpler answer, too.

    Note to self: always check your answer!

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    posted by Steve

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