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

2. Solve e^3x=15

Cheers

**0**

# Log questions - Help

Started by jambo david, Mar 28 2006 07:15 PM

10 replies to this topic

### #1

Posted 28 March 2006 - 07:15 PM

### #2

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:

log

k=10

2. Solve e^3x=15

Answer:

You need a calculator for this..

e

log

3x(log

you should know that log

therefore..

3x =log

x =(log

=0.9 hopefully

Answer:

log

_{10}k=2.3k=10

^{2.3}2. Solve e^3x=15

Answer:

You need a calculator for this..

e

^{3x}=15log

_{e}e^{3x}=log_{e}153x(log

_{e}e) =log_{e}15you should know that log

_{e}e =1therefore..

3x =log

_{e}15x =(log

_{e}15) / 3 (use your calculator )=0.9 hopefully

### #3

Posted 28 March 2006 - 07:40 PM

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

2. Solve e^3x=15

Cheers

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

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

Use the laws of logarithms with powers to get:

ln(e) cancels to get 1 so

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.

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### #4

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 ?

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

### #5

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):

The outcome of which is:

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:

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### #6

Posted 28 March 2006 - 07:46 PM

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

2. Solve e^3x=15

Cheers

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

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

Use the laws of logarithms with powers to get:

ln(e) cancels to get 1 so

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

Posted 28 March 2006 - 07:48 PM

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

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:

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:

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### #8

Posted 28 March 2006 - 07:55 PM

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

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:

the base isn't given

i really appreciate the help

### #9

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.

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### #10

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

log

1

1 =10

log

log

log

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!

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

log

_{1}10^{k}=2.31

^{2.3}=10^{k}1 =10

^{k}log

_{10}1 = log_{10}10^{k}log

_{10}1 = k (log_{10}10)log

_{10}1 = kk = 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

Posted 28 March 2006 - 09:10 PM

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 or .

In fact, I don't think you can even have a logarithm with base 1! Here's why: 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 log

_{1}, you should ask them about it (and then explain it to me )

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

log

_{1}10

^{k}=2.3

1

^{2.3}=10

^{k}

1 =10

^{k}

log

_{10}1 = log

_{10}10

^{k}

log

_{10}1 = k (log

_{10}10)

log

_{10}1 = 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

If you try k=0, you get .

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

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