# Find all numbers x

• Oct 5th 2012, 11:55 AM
TheCracker
Find all numbers x
Find all numbers x that satisfy the given equation

Round your first answer to 5 decimal places and the second one to 1 decimal place.

(log(9x))log(x) = 5

(log(9) + log(x))log(x) = 5

y = log(x)

y^2 + log(9)y -5 = 0

y^2 + 0.95424250943y - 5 = 0

--> y = 1.8092830602959769 or y = -2.763525569725977

--> x = 10^1.8092830602959769 or x = 10^-2.763525569725977

--> x = 64.4589252839 or x = 0.00172375059

So the answer would be

x = 0.00172
x = 64.5

Am I correct?
I tried to solve it that way 3 times already and it said incorrect with other values :/
• Oct 5th 2012, 12:05 PM
HallsofIvy
Re: Find all numbers x
The second answer is not rounded to the number of decimal places the problem requires.

Also, is it possible that the "log" here is supposed to be the natural logarithm, base e, rather than the common logarithm, base 10?
• Oct 5th 2012, 12:11 PM
TheCracker
Re: Find all numbers x
Didn't it say to round the second answer to 1 decimal place? Which is what I did...

And no, it isn't ln
• Oct 5th 2012, 05:05 PM
Prove It
Re: Find all numbers x
You have probably brought in roundoff error by rounding the logarithm BEFORE calculating y. Leave it as \displaystyle \displaystyle \begin{align*} y^2 + \left( \log{9} \right) y - 5 = 0 \end{align*} and apply the Quadratic Formula.
• Oct 5th 2012, 05:19 PM
johnsomeone
Re: Find all numbers x
Quote:

Originally Posted by TheCracker
...
x = 0.00172
x = 64.5

Am I correct?

Up to roundoff error, you should be able to answer that yourself. Plug your solutions back into the original problem and check it.
Your approach (I didn't check your calculations) to solving for x is certainly correct. If you made a mistake, it was a computational one.