Logs and their laws

**David Green** · October 2nd 2011, 12:35 PM

I have a log question;

log 9 x = log 5 (base9) + log 6 (base9) - log 6 (base9)

I have understood it to this point;

log 9 x = log 9 {30/3}

log 9 x = log 9 (10)

I now require to get the "x" on its own, am I going in the right direction from this point as follows;

x = log 9 (10) = log 9^10 = 1000 000 000

I am not to sure what to do with the two log 9's but I require to find the value of "x", am I approaching this correctly?

**skeeter** · October 2nd 2011, 12:41 PM

Originally Posted by David Green

I have a log question;

log 9 x = log 5 (base9) + log 6 (base9) - log 6 (base9)

log 9 x = log 5 (base9)

... the last two terms cancel to 0

...

**David Green** · October 2nd 2011, 12:51 PM

Originally Posted by skeeter

...

I am sorry Skeeter I pushed the wrong key on the last log?

log9 x = log9 5 + log9 6 - log9 3

log9 x = log9 {30/3}

log9 x = log9 (10)

I require to get "x" on its own but have log9 on each side of the equals?

x = log 9^10 = 1000 000 000

Not sure if this is correct?

**skeeter** · October 2nd 2011, 12:57 PM

if $\log_b{x} = \log_b{y}$ , then $x = y$

**David Green** · October 2nd 2011, 01:03 PM

I agree but that does not solve for x?

**Plato** · October 2nd 2011, 01:10 PM

Originally Posted by David Green

I agree but that does not solve for x?

@David Green, why not use basic LaTeX tags?
[TEX]\log_{9}(5)[/TEX] gives $\log_{9}(5)$ .

**David Green** · October 2nd 2011, 01:17 PM

Originally Posted by Plato

@David Green, why not use basic LaTeX tags?
[TEX]\log_{9}(5)[/TEX] gives $\log_{9}(5)$ .

I was told that they are difficult to use and not had any experience with them!

From above it looks like you are saying that "x" = 5 am I correct?

**e^(i*pi)** · October 2nd 2011, 01:41 PM

Originally Posted by David Green

I was told that they are difficult to use and not had any experience with them!

From above it looks like you are saying that "x" = 5 am I correct?

You have $\log_9(10) = \log_9(x)$ which is the same as $x = 10$ (see post 4)

**David Green** · October 3rd 2011, 10:15 AM

Originally Posted by e^(i*pi)

You have $\log_9(10) = \log_9(x)$ which is the same as $x = 10$ (see post 4)

I am still unsure but;

log (base9) x = Log (base9) 10

so I need to get "x" on it's own, therefore because I have no practical examples of how this is done, I am doing it right this way;

Log 10 = 2

**e^(i*pi)** · October 3rd 2011, 10:42 AM

Originally Posted by David Green

I am still unsure but;

log (base9) x = Log (base9) 10

so I need to get "x" on it's own, therefore because I have no practical examples of how this is done, I am doing it right this way;

Log 10 = 2

I'm not sure where you get 2 from. If you like you can raise 9 to the power of both sides - this will have the effect of cancelling out the log function

**David Green** · October 3rd 2011, 11:25 AM

Originally Posted by e^(i*pi)

I'm not sure where you get 2 from. If you like you can raise 9 to the power of both sides - this will have the effect of cancelling out the log function

Log 10 = 2

This is because the two represents an exponent, therefore 10^2 = 100, although I am not confident because as you say my base is 9, therefore;

log 10 = 9

This would be equal to 1 billion as I have posted previously?

**e^(i*pi)** · October 3rd 2011, 11:45 AM

Originally Posted by David Green

Log 10 = 2

This is because the two represents an exponent, therefore 10^2 = 100, although I am not confident because as you say my base is 9, therefore;

log 10 = 9

This would be equal to 1 billion as I have posted previously?

The definition of a logarithm is: $\log_c(b) = a \Leftrightarrow b = c^a$

If we take the expression for a in the first form and put it into the second we get to: $b = c^{\log_c(b)}$

Hence if you raise the base of the logarithm to the log expression they "cancel" in the same way that squaring and square root do.

$9^{\log_9(10)} = 9^{\log_9(x)$

**David Green** · October 3rd 2011, 12:41 PM

Originally Posted by e^(i*pi)

The definition of a logarithm is: $\log_c(b) = a \Leftrightarrow b = c^a$

If we take the expression for a in the first form and put it into the second we get to: $b = c^{\log_c(b)}$

Hence if you raise the base of the logarithm to the log expression they "cancel" in the same way that squaring and square root do.

$9^{\log_9(10)} = 9^{\log_9(x)$

I am not sure I follow the above, however;

log(base9) x = log(base9) 10

if I subtract log(base9) from both sides;

log(base9) - log(base9) x = log(base9) - log(base9) 10

x = 10

Not sure if this is mathematically correct for logs, but it shows x = 10

**mr fantastic** · October 3rd 2011, 08:17 PM

Originally Posted by David Green

I am not sure I follow the above, however;

log(base9) x = log(base9) 10

if I subtract log(base9) from both sides;

log(base9) - log(base9) x = log(base9) - log(base9) 10

x = 10

Not sure if this is mathematically correct for logs, but it shows x = 10

Please pay attention to what you are told or people will just think their time is being wasted.

You have $\log_9 (x) = \log_9 (10)$ .

Post #4 and post #8 tell you excatly what to do. I cannot see why this thread has had to go past post #8.

Thread closed.

**mr fantastic** · October 3rd 2011, 08:22 PM

Originally Posted by Plato

@David Green, why not use basic LaTeX tags?

[TEX]\log_{9}(5)[/TEX] gives $\log_{9}(5)$ .

I was told that they are difficult to use and not had any experience with them!
[snip]

What rot. Did you take the time to check the veracity of what you were "told"? 15 minutes (less time than has been spent on this thread since post #8) is all it takes to learn how to do most of what you need to do: http://www.mathhelpforum.com/math-help/f47/

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