# Thread: Square root in a logarithmic equation

1. ## Square root in a logarithmic equation

Hello,

I am having difficulty isolating x with the square root involved in this equation. I combine the left side with the property of logs that subtraction is equivalent to division, then make both sides an exponent with base 10 to eliminate the log. I then cross multiplied 7x/(9+sqrt(x)) and 100/1. I am then stuck with 900+100(sqrt(x))=7x.

Any help would be greatly appreciated.

Thank you!

2. ## Re: Square root in a logarithmic equation

Originally Posted by mathguy5000

base 10 to eliminate the log. I then cross multiplied 7x/(9+sqrt(x)) and 100/1. I am then stuck with 900+100(sqrt(x))=7x.
\begin{align*}7x&=9\cdot10^2+10^2\sqrt{x} \\(7x- 9\cdot10^2)^2&=10^4x \end{align*}

3. ## Re: Square root in a logarithmic equation

Originally Posted by mathguy5000
I am then stuck with 900+100(sqrt(x))=7x.
sqrt(x) = (7x - 900) / 100

Solve for x by squaring both sides.

4. ## Re: Square root in a logarithmic equation

Ahh. Thank you very much!

5. ## Re: Square root in a logarithmic equation

Does this then become a quadratic equation? I have the final answer in a key, as this is a practice question for my class, but I have tried time and time again and cannot get it to work through to match the solution for the numerical value of x.

6. ## Re: Square root in a logarithmic equation

Originally Posted by mathguy5000
Does this then become a quadratic equation? I have the final answer in a key, as this is a practice question for my class, but I have tried time and time again and cannot get it to work through to match the solution for the numerical value of x.
HERE is a solution.

7. ## Re: Square root in a logarithmic equation

$900+100\sqrt{x}=7x$

Squaring both sides

$(900+100\sqrt{x})^2=(7x)^2$

$810000+10000x^2+180000x=49x^2$

$9951x^2+180000x+810000=0$

$x=\dfrac{-180000\pm\sqrt{(180000)^2-4\times9951\times810000}}{2\times9951}$

$x=-\dfrac{300}{31}$

$x=-\dfrac{900}{107}$

8. ## Re: Square root in a logarithmic equation

Originally Posted by deesuwalka
$900+100\sqrt{x}=7x$

Squaring both sides

$(900+100\sqrt{x})^2=(7x)^2$
OK so far so ...

Originally Posted by deesuwalka
$810000+10000x^2+180000x=49x^2$
Nope - this should be:

$81000 + 10000x +180000\sqrt x = 49 x^2$

which is not much help. And of course your final answers are both negative, which can't be correct.

Better to do as Plato and DenisB have suggested.

To Mathguy500: the quadratic that you need to solve comes from squaring $\displaystyle 7x-900 = 100\sqrt x$, to get:

$\displaystyle 49x^2 - 12600 x +810000 = 10000x$

$\displaystyle 49x^2 - 22600 x +81000 = 0$

Now you can apply the quadratic equation. You'll get two results - you'll have to try both back in the original equation to see which one is correct.

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