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Math Help - Logarithmic expression

  1. #1
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    Logarithmic expression

    Hello guys!

    Could someone help me with this problem? Coz Ive been trying to solve this already for an hour but still I dont get it.

    Heres the question, FInd the exact value of the given expression:

    25^log58

    Thanks in advanced.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by cjru View Post
    Hello guys!

    Could someone help me with this problem? Coz Ive been trying to solve this already for an hour but still I dont get it.

    Heres the question, FInd the exact value of the given expression:

    25^log58

    Thanks in advanced.
    Use b^{\log_bx}=x

    25^{\log_58}=5^{2\log_58}==5^{\log_58^2}=8^2=64
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  3. #3
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    Thank you so much.

    Can you explain your step pls...

    what happened to the 5^log5

    Did it cancelled out?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by cjru View Post
    Thank you so much.

    Can you explain your step pls...

    what happened to the 5^log5

    Did it cancelled out?
    Since the exponential function f(x)=b^x and the logarithmic function g(x)=\log_bx are inverses of each other, their composites are the identity function. That is,

    f[g(x)]=x \ \ and \ \ g[f(x)]=x

    f(\log_bx)=x \ \ and \ \ g(b^x)=x

    b^{\log_bx}=x \ \ and \ \ \log_bb^x=x

    Thus, if their bases are the same, exponential and logarithmic functions "undo" each other. You can use this inverse property of exponents and logarithms to simplify expression like yours.

    25^{\log_58}=5^{(2)\log_58}

    Using (p)\log_bm=\log_bm^p, we get

    5^{\log_58^2}=8^2=64

    I hope I've cleared up any confusion you may have had.
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  5. #5
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    wow! Thats makes the solution clear. Thank you so much. With your explanation I can now solve other related problems in log.

    Thank you.
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