Are the Logs or ln logs

**David Green** · November 18th 2011, 07:23 AM

I have a log question and was wondering which is the correct way to find the value of;

-x being part of the exponent.

log 4^3 - x = log 7

I thought I could find the value of -x this way;

log4^3-x = log7

bring down the -x

xlog64 = log7

divide both sides by log64

x = log7 divided by log64

x = 0.47

Is this permissible?

Thanks

David

**Siron** · November 18th 2011, 07:34 AM

I guess the equation is:
$\log(4^{3-x})=\log(7)$
You can do the following step:
$(3-x)\log(4)=\log7$
$\Leftrightarrow ...$
Or you can solve it like:
$\log(4^{3-x})=\log(7)$
$\Leftrightarrow \log(4^3\cdot 4^{-x})=\log(7)$
$\Leftrightarrow \log(4^3)+\log(4^{-x})=\log(7)$
$\Leftrightarrow ...$

Do you notice the difference with your solution (which is incorrect)? You can't just put -x before the log without doing the same for the 3, you can put (3-x) before the logarithm or you can split the power.

**David Green** · November 18th 2011, 07:40 AM

Originally Posted by Siron

I guess the equation is:
$\log(4^{3-x})=\log(7)$
You can do the following step:
$(3-x)\log(4)=\log7$
$\Leftrightarrow ...$
Or you can solve it like:
$\log(4^{3-x})=\log(7)$
$\Leftrightarrow \log(4^3\cdot 4^{-x})=\log(7)$
$\Leftrightarrow \log(4^3)+\log(4^{-x})=\log(7)$
$\Leftrightarrow ...$

Do you notice the difference with your solution (which is incorrect)? You can't just put -x before the log without doing the same for the 3, you can put (3-x) before the logarithm or you can split the power.

Yes I have noticed that reading through your thread.

Can I ask this because I have seen many posters use the DOUBLE SIDED ARROW in your post, does that mean SAME AS?

David

**Siron** · November 18th 2011, 07:47 AM

It means equivalent to ... take for example:
$\ln(e)=1 \Leftrightarrow e^{1}=e$
It's an equivalence, because
If $\ln(e)=1$ then $e^{1}=e$ , worded in mathematical symbols: $\ln(e)=1 \Rightarrow e^1=e$ , but also
If $e^1=e$ then $\ln(e)=1$ , worded in mathematical symbols: $\ln(e)=1 \Leftarrow e^1=e$
So:
$\ln(e)=1 \Leftrightarrow e^1=e$

But can you solve the equation now? What's your solution now?

**David Green** · November 18th 2011, 07:52 AM

Originally Posted by Siron

It means equivalent to ... take for example:
$\ln(e)=1 \Leftrightarrow e^{1}=e$
It's an equivalence, because
If $\ln(e)=1$ then $e^{1}=e$ , worded in mathematical symbols: $\ln(e)=1 \Rightarrow e^1=e$ , but also
If $e^1=e$ then $\ln(e)=1$ , worded in mathematical symbols: $\ln(e)=1 \Leftarrow e^1=e$
So:
$\ln(e)=1 \Leftrightarrow e^1=e$

But can you solve the equation now? What's your solution now?

In a word NO, I can't see now how to seperate the -x from your method as you have combined it.

**Siron** · November 18th 2011, 08:04 AM

To solve:
$\log(4^{3-x})=\log(7)$
$\Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)$
$\Leftrightarrow \log(64)+\log(4^{-x})=\log(7)$
$\Leftrightarrow \log(64)-x\log(4)=\log(7)$

Other way:
$\Leftrightarrow (3-x)\log(4)=\log(7)$
$\Leftrightarrow 3\log(4)-x\log(4)=\log(7)$

Is this clear?

**Plato** · November 18th 2011, 08:06 AM

Originally Posted by David Green

In a word NO, I can't see now how to seperate the -x from your method as you have combined it.

Mr. Green, you are making too much out of this.
$\log(4^{3-x})=(3-x)\log(4)$

$(3-x)=\frac{\log(7)}{\log(4)}$

$x=\left(3-\frac{\log(7)}{\log(4)}\right)$

**David Green** · November 18th 2011, 08:12 AM

Originally Posted by Siron

To solve:
$\log(4^{3-x})=\log(7)$
$\Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)$
$\Leftrightarrow \log(64)+\log(4^{-x})=\log(7)$
$\Leftrightarrow \log(64)-x\log(4)=\log(7)$

Other way:
$\Leftrightarrow (3-x)\log(4)=\log(7)$
$\Leftrightarrow 3\log(4)-x\log(4)=\log(7)$

Is this clear?

In a word NO, there is an overwhelming amount of information there too much to take in. I think it's best to possibly stick with your second example and then I can learn how to find the value of X from that point onwards.

Thanks

David

**David Green** · November 18th 2011, 08:19 AM

Originally Posted by Plato

Mr. Green, you are making too much out of this.
$\log(4^{3-x})=(3-x)\log(4)$

$(3-x)=\frac{\log(7)}{\log(4)}$

$x=\left(3-\frac{\log(7)}{\log(4)}\right)$

Plato, thank you this method is much clearer. Looking back on the other thread had too much detail and I could not get my head round how to seperate the X from the logs, you made this very clear.

David

**Siron** · November 18th 2011, 10:47 AM

Originally Posted by David Green

In a word NO, there is an overwhelming amount of information there too much to take in. I think it's best to possibly stick with your second example and then I can learn how to find the value of X from that point onwards.

Thanks

David

Can you say which step you don't understand etc ... ? (that's important!) Actually it's just basic algebra.

**David Green** · November 18th 2011, 01:13 PM

Originally Posted by Siron

To solve:
$\log(4^{3-x})=\log(7)$
$\Leftrightarrow \log(4^{3}\cdot 4^{-x})=\log(7)$
$\Leftrightarrow \log(64)+\log(4^{-x})=\log(7)$
$\Leftrightarrow \log(64)-x\log(4)=\log(7)$

Other way:
$\Leftrightarrow (3-x)\log(4)=\log(7)$
$\Leftrightarrow 3\log(4)-x\log(4)=\log(7)$

Is this clear?

Your last thread asked me to explain the part that is not clear?

If you look at what you posted and try to look at it from a view that somebody doesn't know how to get x on it's own, then look where x is in the logs you could forgive for not knowing how to get it on it's own?

Looking now at the second example you gave;

(3 - x)log(4) = log(7), here I would divide the LHS to get log(4) on the RHS

(3 - x)log(4) / log(4) = log(7) / log(4), this leaves;

3 - x = log(7) / log(4), from here and only because I have seen it from PLATO thread I know to move the 3 to the RHS and change the sign;

x = 3 - log(7) / log(4)

x = 1.596

x = 1.60 (3.s.f) although I have a little doubt about the 0 because 0 is not normally significant?

**Siron** · November 19th 2011, 01:29 AM

Originally Posted by David Green

If you look at what you posted and try to look at it from a view that somebody doesn't know how to get x on it's own, then look where x is in the logs you could forgive for not knowing how to get it on it's own?

Offcourse, but I (and the others) don't know what your knowledge is about algebra so it's difficult to estimate what you already have learned, not learned yet, etc.
But I think you understand it now.

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