Logarithmic Troubles

**Jbuckham** · April 6th 2009, 09:44 PM

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?

**Reckoner** · April 6th 2009, 10:05 PM

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.

$1000e^{1.3y}=2000$

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

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

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

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

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

**Jbuckham** · April 6th 2009, 10:51 PM

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...

**Grandad** · April 6th 2009, 11:05 PM

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?)

Grandad

**Reckoner** · April 6th 2009, 11:13 PM

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.

**Jbuckham** · April 6th 2009, 11:33 PM

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

**stapel** · April 7th 2009, 03:36 AM

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.

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