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Log questions - Help


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#1 jambo david

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Posted 28 March 2006 - 07:15 PM

1. Given that log10^k=2.3, write down an expression for the exact value of k

2. Solve e^3x=15

Cheers sad.gif

#2 SncZ

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Posted 28 March 2006 - 07:34 PM

1. Given that log10^k=2.3, write down an expression for the exact value of k

Answer:
log10k=2.3
k=102.3


2. Solve e^3x=15

Answer:

You need a calculator for this..

e3x =15
logee3x =loge15
3x(logee) =loge15

you should know that logee =1

therefore..

3x =loge15
x =(loge15) / 3 (use your calculator wink.gif )

=0.9 hopefully tongue.gif

#3 The Wedge Effect

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Posted 28 March 2006 - 07:40 PM

QUOTE(jambo david @ Mar 28 2006, 08:15 PM) View Post

1. Given that log10^k=2.3, write down an expression for the exact value of k

2. Solve e^3x=15

Cheers sad.gif


I'm a bit unclear on the first question. Is it asking you to find 

\begin{align*}\log_{10}^k=2.3\,.\end{align*}

or 

\begin{align*}\log_{10} k=2.3\,.\end{align*}

?

To find the answer for question two, you take logs of both side using the natural logarithm.



\begin{align*}ln.e^{3x} = ln.15\end{align*}

Use the laws of logarithms with powers to get:



\begin{align*} 3x.ln.e = ln.15\\ .\end{align*}

ln(e) cancels to get 1 so



\begin{align*} 3x=ln.15 \\.\end{align*}



\begin{align*}x=\frac {ln 15}{3}\\ \,.\end{align*}

In reply to SncZ, my answer would be sufficient, it's not that necessary to use a calculator, you can express it in simplified form like this, unless it specifically asks for the value to a number of decimal places.

#4 SncZ

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Posted 28 March 2006 - 07:41 PM

wedge how do you get the 'big divide sign'

and for natural log did you just type 'ln' or is there a code for the symbol ?

#5 The Wedge Effect

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Posted 28 March 2006 - 07:44 PM

If you hover over the expressions, you will see the LaTeX code I used for the expressions. I just typed in 'ln' instead of using a special code, because I wasn't sure if one existed for the natural logarithm.

For the divide sign, I used the following code (example of an expression with the code is used):

CODE
[display]f(x)=\frac {x^2+2x+1}{x^2-2x+1}[/display]


The outcome of which is:



\begin{align*}f(x)=\frac {x^2+2x+1}{x^2-2x+1}\end{align*}

#6 jambo david

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Posted 28 March 2006 - 07:46 PM

QUOTE(The Wedge Effect @ Mar 28 2006, 08:40 PM) View Post

QUOTE(jambo david @ Mar 28 2006, 08:15 PM) View Post

1. Given that log10^k=2.3, write down an expression for the exact value of k

2. Solve e^3x=15

Cheers sad.gif


I'm a bit unclear on the first question. Is it asking you to find 

\begin{align*}\log_{10}^k=2.3\,.\end{align*}

or 

\begin{align*}\log_{10} k=2.3\,.\end{align*}

?

To find the answer for question two, you take logs of both side using the natural logarithm.



\begin{align*}ln.e^{3x} = ln.15\end{align*}

Use the laws of logarithms with powers to get:



\begin{align*} 3x.ln.e = ln.15\\ .\end{align*}

ln(e) cancels to get 1 so



\begin{align*} 3x=ln.15 \\.\end{align*}



\begin{align*}x=\frac {ln 15}{3}\\ \,.\end{align*}

In reply to SncZ, my answer would be sufficient, it's not that necessary to use a calculator, you can express it in simplified form like this, unless it specifically asks for the value to a number of decimal places.



question 1 is Log10^k=2.3

the 10 is big if you know what i mean, normal sized

#7 The Wedge Effect

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Posted 28 March 2006 - 07:48 PM

Ah, in that case, you use the laws of logarithms to simplify the expression:



\begin{align*}\log{10^k}=2.3\end{align*}


\begin{align*}k\log{10}=2.3\end{align*}

Bear with me, I need to work on this a bit. Is the base 10 or something else (if one is given)?

Edit: If no bases are given, I'll assume the answer is:



\begin{align*} k=\frac{2.3}{\log{10}}\end{align*}

#8 jambo david

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Posted 28 March 2006 - 07:55 PM

QUOTE(The Wedge Effect @ Mar 28 2006, 08:48 PM) View Post

Ah, in that case, you use the laws of logarithms to simplify the expression:



\begin{align*}\log{10^k}=2.3\end{align*}


\begin{align*}k\log{10}=2.3\end{align*}

Bear with me, I need to work on this a bit. Is the base 10 or something else (if one is given)?

Edit: If no bases are given, I'll assume the answer is:



\begin{align*} k=\frac{2.3}{\log{10}}\end{align*}

the base isn't given

i really appreciate the help

#9 The Wedge Effect

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Posted 28 March 2006 - 07:56 PM

You're welcome. I wouldn't rely on my answer for k though, I'm not sure if it's correct. rolleyes.gif

#10 SncZ

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Posted 28 March 2006 - 08:02 PM

thanks wedge,

for number 1, if there is no 'to the base of' we just got told to assume it's 1

log110k =2.3

12.3 =10k
1 =10k
log101 = log1010k
log101 = k (log1010)
log101 = k
k = 0


im not sure if thats right.. have a feeling its not, but its the best i can do at the moment



thats what i was told anyway (about the bases), im not trying to say wedge is wrong!

#11 George

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Posted 28 March 2006 - 09:10 PM

QUOTE(SncZ @ Mar 28 2006, 09:02 PM) View Post

for number 1, if there is no 'to the base of' we just got told to assume it's 1

That seems really odd - usually "log" is taken to mean either \log _{10} or \log _e.

In fact, I don't think you can even have a logarithm with base 1! Here's why: \log _1 x = y means y is the power of 1 that gives x. Because 1 to any power is 1, you'll have trouble finding y if x isn't 1!

If your teacher has really told you to use log1, you should ask them about it (and then explain it to me biggrin.gif )


QUOTE(SncZ @ Mar 28 2006, 09:02 PM) View Post

for number 1, if there is no 'to the base of' we just got told to assume it's 1

log110k =2.3

12.3 =10k
1 =10k
log101 = log1010k
log101 = k (log1010)
log101 = k
k = 0

im not sure if thats right.. have a feeling its not, but its the best i can do at the moment

thats what i was told anyway (about the bases), im not trying to say wedge is wrong!

Your feeling is right, unfortunately smile.gif

If you try k=0, you get \log 10^0 = \log 1 = 0 \neq 2.3.

Wedge's working is certainly correct, and it's as far as you can really take it.





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