1. ## Logarithmic Troubles

The Equation is $x(y) = 1000e^{1.3y}$ x = 2000 find y
So applied Logarithms to both sides which left me:

$
\log_{10} 2000 + \log_{10} y = \log_{10} 1000 + 1.3y \log_{10} e
$

$
3.30103 + \log_{10} y= 3 + 1.3y(0.43429)
$

$
\log_{10} y = 3 + 0.56458y -3.30103
$

$
\log_{10} y = 0.56458y -0.30103
$

I'm not sure where to go from here?

2. Originally Posted by Jbuckham
The Equation is $x(y) = 1000e^{1.3y}$ x = 2000 find y
So applied Logarithms to both sides which left me:

$
\log_{10} 2000 + \color{red}\log_{10} y\color{black} = \log_{10} 1000 + 1.3y \log_{10} e
$
Where did that come from? Proceed instead as follows.

$1000e^{1.3y}=2000$

$\Rightarrow\ln\left(1000e^{1.3y}\right)=\ln2000$

$\Rightarrow\ln1000+\ln e^{1.3y}=\ln2000$

$\Rightarrow1.3y=\ln2000-\ln1000$

$\Rightarrow1.3y=\ln\frac{2000}{1000}$

$\Rightarrow y=\frac{10}{13}\ln2$

3. Originally Posted by Reckoner
Where did that come from? Proceed instead as follows.

The original equation is $x \color{red}\ (y) \color{black} = 1000e^{1.3y}$
therefore $Log x + Log y = log 1000 + 1.3ylog e$ or $1000e^{1.3y}=2000y$
Thats where that log y came from...

4. Hello Jbuckham
Originally Posted by Jbuckham
The original equation is $x \color{red}\ (y) \color{black} = 1000e^{1.3y}$
therefore $Log x + Log y = log 1000 + 1.3ylog e$ or $1000e^{1.3y}=2000y$
Thats where that log y came from...
I think it's a simple case of not interpreting the notation correctly.

You have interpreted $x(y)$ to mean $x \times y$. But I think that, in this question, $x(y) = ...$ is function notation, and means $x$, expressed as a function of $y$, is equal to ...

So the equation you're being asked to solve is simply:

$x = 1000e^{1.3y}$

(Why do people make life more difficult than it needs to be, eh?)

5. Originally Posted by Jbuckham
The original equation is $x \color{red}\(y)\color{black} = 1000e^{1.3y}$
therefore $Log x + Log y = log 1000 + 1.3ylog e$
Thats where that log y came from...
Oh, I thought when you wrote $x(y)$ that you were specifying $x$ as a function of $y$. Instead, we have

$1000e^{1.3y}=2000y,$

which has no solutions. You can see from their graphs: the exponential function starts out at much higher values than the linear function, and then it begins to grow at such a quickly increasing rate that the linear function can never hope to catch up.

Edit: Grandad may be right--perhaps you have misinterpreted the problem? You may want to state the question exactly as it is written in your book or assignment.

6. You're right I have misinterpreted the problem. Thank you very much for your help. it had me completly stumpted

7. Originally Posted by Jbuckham
The original equation is $x \color{red}\ (y) \color{black} = 1000e^{1.3y}$
therefore $Log x + Log y = log 1000 + 1.3ylog e$...
Those parentheses on the left-hand side of the "equals" sign do not indicate multiplication!

To learn how to work with "function notation", try here.