# [SOLVED] Log Function

• Feb 10th 2009, 05:26 AM
MathLoser
[SOLVED] Log Function
Real variables:

ln(2+y) = 5ln(3-x) -2sqrtx

(where sqrtx = the square root of x sorry!)

I need to find the range of values of x and y for which they're defined and express as y = f(x)

Can someone show me how to do these types of questions so I can do the rest? I'm stuck!!

Many thanks, I appreciate it (Headbang)
• Feb 10th 2009, 06:20 AM
Amanda H
First of all to express as $\displaystyle y = f(x)$ take the exponential of both sides and then subtract 2 from both sides:

$\displaystyle \ln(2+y) = 5\ln(3-x) - 2\sqrt{x}$
$\displaystyle 2+y = e^{5\ln(3-x) - 2\sqrt{x}}$
$\displaystyle y= e^{5\ln(3-x) - 2\sqrt{x}} - 2$

Then for the values for which x and y are defined you have to look at which values are not possible in the original equation:
$\displaystyle \ln(2+y) = 5\ln(3-x) - 2\sqrt{x}$

Dealing with values of x and first looking at the square root sign it must be true that $\displaystyle x\geq0$ otherwise we would be taking the square root of a negative number which isn't allowed. Going on to look at $\displaystyle \ln(3-x)$ we must use values $\displaystyle x\leq 3$ or else we would be taking the ln of a number $\displaystyle >0$ which does not exist.
So the values of x for which the equation is defined would be:

$\displaystyle 0 \leq x \leq 3$ where x is a member of $\displaystyle \Re$

Looking at values for y there is only one restriction shown by $\displaystyle \ln(2+y)$ in the same way to the earlier log it must be $\displaystyle \geq 0$ so:

$\displaystyle y\geq -2$ where y is a member of $\displaystyle \Re$

If you have further questions just ask.
• Feb 10th 2009, 06:35 AM
Soroban
Hello, MathLoser!

This is a tricky one . . .

Quote:

$\displaystyle \ln(2+y) \;= \;5\ln(3-x) -2\sqrt{x}$

Find the range of values of $\displaystyle x$ and $\displaystyle y$ for which they're defined
and express as: $\displaystyle y \,=\,f(x)$

From $\displaystyle \sqrt{x}$, we see that: $\displaystyle x \geq 0$

From $\displaystyle \ln(3-x)$, we see that: $\displaystyle x < 3$

. . The domain is: .$\displaystyle \boxed{\;0\: \leq\: x\: <\: 3\;}$

We have: .$\displaystyle \ln(2+y) \:=\:\ln(3-x)^5 - 2\sqrt{x}$

Then: .$\displaystyle y + 2 \;=\;e^{\ln(3-x)^5 - 2\sqrt{x}} \;=\;e^{\ln(3-x)^5}\cdot e^{-2\sqrt{x}} \;=\;\frac{e^{\ln(3-x)^5}}{e^{2\sqrt{x}}} \;=\;\frac{(3-x)^5}{e^{2\sqrt{x}}}$

. . Hence: .$\displaystyle \boxed{y \;=\;\frac{(3-x)^5}{e^{2\sqrt{x}}}-2}$

When $\displaystyle x = 0\!:\;y \:=\:\frac{3^5}{e^0} - 2 \:=\:241$

When $\displaystyle x = 1\!:\;\;y \:=\:\frac{2^5}{e^2} \:\approx\:2.331$

When $\displaystyle x = 2\!:\;\;y \:=\:\frac{1^5}{2^{2\sqrt{2}}} \:\approx\:-1.982$

When $\displaystyle x\to3\!:\;\;y \to -2$

The range seems to be: .$\displaystyle \boxed{\;-2 \:<\: y \:\leq \:241\;}$

• Feb 10th 2009, 06:49 AM
MathLoser
Hi and thanks to both of you for your posts - do they agree or are there slightly different answers? :o