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Math Help - [SOLVED] Proving log problem

  1. #1
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    [SOLVED] Proving log problem

    Show that:

    \log_{x^2}(y) = \log_x\sqrt{y} \

     x > 0 and y > 1

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  2. #2
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    \log_{x^2}y=\frac{\log_xy}{\log_xx^2}=\frac{\log_x  y}{2}=\frac{1}{2}\log_xy=\log_x(y)^{\frac{1}{2}}=\  log_x\sqrt{y}
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  3. #3
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    Quote Originally Posted by MarcoMP View Post
    Show that:

    \log_{x^2}(y) = \log_x\sqrt{y} \

     x > 0 and y > 1

    Use the base change formula:

    \log_a(x)=\dfrac{\ln(x)}{\ln(a)}

    With your example you have:

    \begin{aligned}\log_{x^2}(y)&= \\<br />
&= \dfrac{\ln(y)}{\ln(x^2)} \\<br />
&=\dfrac{\ln(y)}{2\ln(x)} \\<br />
&=\dfrac{\frac12 \ln(y)}{\ln(x)} = \dfrac{\ln(\sqrt{y})}{\ln(x)} = \log_x(\sqrt{y})\end{aligned}
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  4. #4
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    Could you please explain why both bases are x when you put in fraction form?

    earboth could you do that using only logs laws seeing as I dont understand the In yet
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  5. #5
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    Quote Originally Posted by MarcoMP View Post
    Could you please explain why both bases are x when you put in fraction form?

    earboth could you do that using only logs laws seeing as I dont understand the In yet
    Have a look at red-dog's solution: He only uses \log_x and he provides you with a much better and elegant solution than I did.
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  6. #6
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    \log_{x^2}y=\frac{\log_{10}{y}}{\log_{10}{x^2}}=\f  rac{\log_{10}y}{2log_{10}x}=\frac{1}{2}\log_xy=\lo  g_x(y)^{\frac{1}{2}}=\log_x\sqrt{y}

    Is that correct?
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  7. #7
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    Quote Originally Posted by MarcoMP View Post
    \log_{x^2}y=\frac{\log_{10}{y}}{\log_{10}{x^2}}=\f  rac{\log_{10}y}{2log_{10}x}=\frac{1}{2}\log_xy=\lo  g_x(y)^{\frac{1}{2}}=\log_x\sqrt{y}

    Is that correct?
    Perfect!
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  8. #8
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    Thanks very much! Now just got to get x in:

    e^{x^2} = \frac{1}{\sqrt {3^{x^5}}} I'm quite lost i must say
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  9. #9
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    Quote Originally Posted by MarcoMP View Post
    Thanks very much! Now just got to get x in:

    e^{x^2} = \frac{1}{\sqrt {3^{x^5}}} I'm quite lost i must say
    1. Please do yourself and do us a favour and start a new thread if you have a new question.

    2. This one looks very ugly to me. Are you sure you typed it correctly here? (I'm only asking because the key "e" is exactly under the key"3" and maybe you slipped a little bit?) If the e is meant to be a 3 then your question becomes:

    3^{x^2} = \frac{1}{\sqrt {3^{x^5}}} ~\implies~ 3^{x^2} \cdot \sqrt {3^{x^5}}= 1~\implies~3^{x^2+\frac52 x^2} = 1=3^0~\implies~\frac72x^2=0~\implies~x=0
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