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Math Help - Trouble

  1. #1
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    Trouble

    Solve the equation

    In10^x=Log e^x +2

    Now I got

    xIn10 = xLog e +2
    x 2.302585093 = x 0.434294481 + 2

    x = 0.434294481 +2/2.302585093

    so x = 1.0572066

    But imputting this into the original equation is incorrect. Where did I go wrong? Help would be much appreciated thank you.

    Simon
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  2. #2
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    Re: Trouble

    Quote Originally Posted by bigtumbler View Post
    Solve the equation

    In10^x=Log e^x +2

    Now I got

    xIn10 = xLog e +2
    x 2.302585093 = x 0.434294481 + 2

    x = 0.434294481 +2/2.302585093

    so x = 1.0572066

    But imputting this into the original equation is incorrect. Where did I go wrong? Help would be much appreciated thank you.

    Simon
    This is almost unreadable. What base is the logarithm on the right hand side? If it's base e, why not just use ln like on the LHS? Assuming it is...

    \displaystyle \begin{align*} \ln{\left(10^x\right)} &= \ln{\left(e^x\right)} + 2 \\ \ln{\left(10^x\right)} - \ln{\left(e^x\right)} &= 2 \\ \ln{\left(\frac{10^x}{e^x}\right)} &= 2 \\ \ln{\left[\left(\frac{10}{e}\right)^x\right]} &= 2 \\ x\ln{\left(\frac{10}{e}\right)} &= 2 \\ x &= \frac{2}{\ln{\left(\frac{10}{e}\right)}} \\ x &= \frac{2}{\ln{10} - \ln{e}} \\ x &= \frac{2}{\ln{10} - 1} \end{align*}
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  3. #3
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    Re: Trouble

    This is how the equation was written to me. I am not sure why you used (In) on the RHS sinc In(e)=1 and log(e^1) = 0.434 approx.

    Simon
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  4. #4
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    Re: Trouble

    \ln 10^x=\log_{10}e^x+2

    x\ln 10=x\log_{10}e+2

    x \cdot \left(\ln 10-\frac{1}{\ln 10}\right)=2

    Hence :

    x=\frac{2\ln 10}{\ln^2 10-1}
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  5. #5
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    Re: Trouble

    Thank you very much . I didn't know log10e = 1/In10. How come you used a minus instead of In10/(1/In10) a division sign? Sorry that my equations are not that clear. Unsure how to format it otherwise.
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  6. #6
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    Re: Trouble

    Quote Originally Posted by bigtumbler View Post
    Thank you very much . I didn't know log10e = 1/In10. How come you used a minus instead of In10/(1/In10) a division sign? Sorry that my equations are not that clear. Unsure how to format it otherwise.
    x\cdot a=x\cdot b+c \Rightarrow x\cdot a-x\cdot b=c\Rightarrow x \cdot(a-b)=c
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  7. #7
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    Re: Trouble

    Ok so if it was for example 6x=4x +2 you would minus it (6x-4x=2) ? I always thought you had to divide because 6x=12 you divide.
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  8. #8
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    Re: Trouble

    Do you not understand why you do those things? Your objective is to solve for x- that is, get x alone on one side. Since, in the second case, the only reason x is NOT alone is because it is multiplied by 6, it makes sense to divide by 6 to get 6x/6= 12/6. But that is NOT the case in "6x= 4x+ 2". You could divide both sides by 6- you would get x= (4/6)x+ (2/6)= (2/3)x+ (1/3). But because you still have x on both sides of the equation, that does not help a lot- and it has those annoying fractions. Instead, it is better to get rid of that "4x" on the right side. And because it is added you get rid of 4x by subtracting it from both sides:
    6x- 4x= 4x+ 2- 4x or 2x= 2. Now what do you do to find x?

    Also, going back to the original problem, you really need to distinguish between "I" (large letter "i") and "l" (small letter "L"). The symbol for the natural logarithm is "ln", not "In".
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