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Math Help - Evaluating and finding x; Logarithms

  1. #1
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    Evaluating and finding x; Logarithms

    a.) so i was trying to evaluate this
    (log5 4 x log2 10)/log25 sqrt10

    so what i did was do change of base on the numerator than i got 4 (is doing this correct?) then after i got this well i got stuck on what to do next. help?

    b.) one last i came across this problem log3 (x+2)=3logx 3
    i got stuck on this as well, what would i do to find x?

    thank you for your time
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  2. #2
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    Re: Evaluating and finding x; Logarithms

    Quote Originally Posted by yety124 View Post
    [snip]

    b.) one last i came across this problem log3 (x+2)=3logx 3
    i got stuck on this as well, what would i do to find x?

    thank you for your time
    The right hand side is \log_x (27). Using the change of base rule, this can be written as \frac{3}{\log_3 (x)} ....
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  3. #3
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    Re: Evaluating and finding x; Logarithms

    Hello, yety124!

    \text{(a) Simplify: }\:\frac{\log_54 \cdot \log_210}{\log_{25}\sqrt{10}}

    \text{We have: }\:\frac{\log_5(4) \cdot \log_2(10)}{\log_{25}(10^{\frac{1}{2}})} .[1]

    . . \log_5(4) \:=\:\log_5(2^2) \:=\:2\log_5(2)

    . . \log_2(10) \:=\:\frac{\log_5(10)}{\log_5(2)}

    . . \log_{25}(10)^{\frac{1}{2}} \:=\:\frac{\log_5(10^{\frac{1}{2}})}{\log_5(25)} \:=\:\frac{\log_5(10^{\frac{1}{2}})}{\log_5(5^2)} \;=\;\frac{\frac{1}{2}\log_5(10)}{2\,\log_5(5)}} \:=\:\frac{\log_5(10)}{4}


    Substitute into [1]:

    . . \frac{2\,\rlap{//////}{\log_5(2)}\cdot\dfrac{\rlap{///////}\log_5(10)}{\rlap{//////}\log_5(2)}} {\dfrac{\rlap{///////}\log_5(10)}{4}} \;=\;8

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    Re: Evaluating and finding x; Logarithms

    ok i get the first one now thanks
    but then the second one tried to solve it this is where i got to
    log3 (x+2) * log3 (x)=3
    what do i do after this?
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    Re: Evaluating and finding x; Logarithms

    Quote Originally Posted by yety124 View Post
    ok i get the first one now thanks
    but then the second one tried to solve it this is where i got to
    log3 (x+2) * log3 (x)=3
    what do i do after this?
    At this point I'd re-check the question, and also check that there was not a typo in the question.
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  6. #6
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    Re: Evaluating and finding x; Logarithms

    HI yety124,
    For the second one
    logB3 (x+2) =3logBx (3)
    log B3 (x+2) =log Bx (27)= logB3 (27)/logB3(x)
    logB3 (x+2)* log B3 (x) = logB3 (27)
    x^2 + 2x = 27
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    Re: Evaluating and finding x; Logarithms

    Quote Originally Posted by bjhopper View Post
    HI yety124,
    For the second one
    logB3 (x+2) =3logBx (3)
    log B3 (x+2) =log Bx (27)= logB3 (27)/logB3(x)
    logB3 (x+2)* log B3 (x) = logB3 (27)
    x^2 + 2x = 27
    The left hand side of the final line follows from an invalid log rule and is wrong. Products of logs do not simplify in the way this line suggests .....

    If the * had been a +, then the last line would be correct.
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