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Thread: Solve log(x+10) for x=10/9

  1. #1
    Junior Member B9766's Avatar
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    Solve log(x+10) for x=10/9

    I'm having trouble with a statement in the text that says, "the equation $log_{10}\ (x +\ 10)\ \neq \ log_{10}\ x\ +\ log_{10}\ 10$ for all values of x. However, you can show that this equation does have a solution: $x\ =\ 10/9$

    I understand the first part of the statement but haven't been able to figure out how they arrived at $x\ =\ 10/9$.

    I tried starting with $f(x)\ = \ log\ (x\ +\ 10)$ and then solving for x. But doing so I get:

    $y\ = \ log(x\ +\ 10)$
    $y^{10}\ =\ (x\ +\ 10)$
    $x\ =\ y^{10}\ -\ 10$ and that's clearly not $10/9$

    I also tried substituting $10/9$ in the first statement:
    $log(10/9\ +\ 10)\ =\ log(\frac{10+90}{9})\ =\ log(100/9)\ =\ log(10^2)\ -\ log(9)\ = 2(1)\ -\ log(9)$ and that's not $10/9$

    I would appreciate help solving this one
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    Re: Solve log(x+10) for x=10/9

    What do you mean this equation has a solution? Please post the exact wording of the question, because what you are asking does not make sense. If you are asked to "evaluate the expression when $x=\dfrac{10}{9}$", that is very different from a "solution" to the equation. You have not shown us an equation. The solution to $f(x) = \log_{10}(x+10)$ is not meaningful. This is an equation that is "solved" by any $x>-10$.
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    Forum Admin topsquark's Avatar
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by B9766 View Post
    I'm having trouble with a statement in the text that says, "the equation $log_{10}\ (x +\ 10)\ \neq \ log_{10}\ x\ +\ log_{10}\ 10$ for all values of x. However, you can show that this equation does have a solution: $x\ =\ 10/9$

    I understand the first part of the statement but haven't been able to figure out how they arrived at $x\ =\ 10/9$.

    I tried starting with $f(x)\ = \ log\ (x\ +\ 10)$ and then solving for x. But doing so I get:

    $y\ = \ log(x\ +\ 10)$
    $y^{10}\ =\ (x\ +\ 10)$
    $x\ =\ y^{10}\ -\ 10$ and that's clearly not $10/9$

    I also tried substituting $10/9$ in the first statement:
    $log(10/9\ +\ 10)\ =\ log(\frac{10+90}{9})\ =\ log(100/9)\ =\ log(10^2)\ -\ log(9)\ = 2(1)\ -\ log(9)$ and that's not $10/9$

    I would appreciate help solving this one
    But $\displaystyle log(10/9 + 10) \neq x$. You wish to show that x = 10/9 gives the same number for both sides, not that they are both equal to x. ie. What is $\displaystyle log(10/9) + log(10)$?

    -Dan
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by topsquark View Post
    But $\displaystyle log(10/9 + 10) \neq x$. You wish to show that x = 10/9 gives the same number for both sides, not that they are both equal to x. ie. What is $\displaystyle log(10/9) + log(10)$?

    -Dan
    If the OP is trying to solve $x=\log_{10}(x+10)$, then $x \approx \dfrac{10}{9}$ is very close to the solution. It is off by about 0.068.
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    Forum Admin topsquark's Avatar
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by SlipEternal View Post
    If the OP is trying to solve $x=\log_{10}(x+10)$, then $x \approx \dfrac{10}{9}$ is very close to the solution. It is off by about 0.068.
    Interesting. I had thought the equation to solve was $\displaystyle log_{10}(x + 10) = log_{10}(x) + log_{10}(10)$ and to find the solution is x = 10/9.

    Perhaps the OP can validate which?

    -Dan
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by topsquark View Post
    Interesting. I had thought the equation to solve was $\displaystyle log_{10}(x + 10) = log_{10}(x) + log_{10}(10)$ and to find the solution is x = 10/9.

    Perhaps the OP can validate which?

    -Dan
    I think what you suggested is correct.
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by SlipEternal View Post
    What do you mean this equation has a solution? Please post the exact wording of the question, because what you are asking does not make sense. If you are asked to "evaluate the expression when $x=\dfrac{10}{9}$", that is very different from a "solution" to the equation. You have not shown us an equation. The solution to $f(x) = \log_{10}(x+10)$ is not meaningful. This is an equation that is "solved" by any $x>-10$.
    EXACTLY as shown in the textbook, it says under Study Tips, "Recall that the logarithm of a sum is not the sum of the logarithms. That is, the equation $log_{10}(x+ 10) ≠ log_{10} x + log_{10} 10$ for all values of x. However, you can show that this equation does have a solution: x = 10/9"
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    Junior Member B9766's Avatar
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    Re: Solve log(x+10) for x=10/9

    I was grasping at straws, topsquark. I obviously don't know how to resolve the statement in the text book. Please see my response to SlipEternal to see the EXACT wording from the text book.
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    Re: Solve log(x+10) for x=10/9

    Please show me the steps that allow you to start with $x\ =\ log_{10}(x\ +\ 10)$ and arrive at $x\ \approx \ \frac{10}{9}$

    I must have been round about but that was my real question.
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    Re: Solve log(x+10) for x=10/9

    Hint:

    $\displaystyle \log_a(b)+\log_a(c)=\log_a(bc)$
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    Re: Solve log(x+10) for x=10/9

    Quote Originally Posted by B9766 View Post
    Please show me the steps that allow you to start with $x\ =\ log_{10}(x\ +\ 10)$ and arrive at $x\ \approx \ \frac{10}{9}$

    I must have been round about but that was my real question.
    If you have $y=\log_a b$ then $a^y=b$. Applying that same logic, we have $x = \log_{10}(x+10)$, so

    $$10^x = x+10$$

    Then you have $10^x-x-10=0$. There is no easy solution to finding solutions to an equation with a sum of an exponential term with a linear one. If we put $f(x) = 10^x-x-10$, we can find where one solution lies by plugging in points: $f(1)=-1$ while $f(2) = 88$. So, there is a solution somewhere between 1 and 2 because $f(x)$ is continuous, so it must achieve every value between $-1$ and $88$, including zero, on the interval $(1,2)$. You need a numerical analysis system to determine the actual value for $x$ that satisfies the equality. When you find that solution, you find that it is fairly close to $\dfrac{10}{9}$. But, this is purely coincidence, and has nothing to do with the question your book asked.

    It wanted to know the solution to: $\log_{10}(x+10) = \log_{10}x+\log_{10}10$. You should know that $\log_a b + \log_a c = \log_a (bc)$. Applying this to the RHS, you have:

    $$\log_{10}(x+10) = \log_{10}(10x)$$

    Taking each side as a power of 10, we have:

    $$x+10 = 10x \Longrightarrow x = \dfrac{10}{9}$$
    Last edited by SlipEternal; Jul 13th 2018 at 11:28 AM.
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    Re: Solve log(x+10) for x=10/9

    I'm aware of the rule. I don't see how it applies.
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  13. #13
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    Re: Solve log(x+10) for x=10/9

    SlipEternal,

    OK. Now it's clear. It really was calling an inequality an "equation". That confused me. If you ignore the point they were making (logs of sums are not the same as sums of logs) and turn the inequality into an equality (which seems like a really stupid way to teaching a concept), then solve for x, you will get the answer.

    Thanks for taking the time to answer this. I appreciate it.
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