1. ## Logarithmic Explaination Needed

Q: Given that Lg x = p and lg y = q, express the following in terms of p and q.
a) xy

Ans ￼: = 10^p x 10^q
= 10^(p+q)

Where the 10 come from?! Like lg10, but if it is it will be xlg10 which is x(1). I don't get this. Please help.

2. ## Re: Logarithmic Explaination Needed

$\log(x)=y \Leftrightarrow 10^{y}=x$

They use the logarithm with base 10 which is standard. If the base wasn't 10 the autors would have indicated that for sure. And it's not useful to write the base 10:

3. ## Re: Logarithmic Explaination Needed

Originally Posted by mushhall
Q: Given that Lg x = p and lg y = q, express the following in terms of p and q.
a) xy

Ans ￼: = 10^p x 10^q
= 10^(p+q)

Where the 10 come from?! Like lg10, but if it is it will be xlg10 which is x(1). I don't get this. Please help.
Whoever wrote this question and answer is using $\text{Lg}(x)$ as most use $\log_{10}(x)$.

So $\text{Lg}(x)=p$ is the same as $x=10^p$.
Also you are correct $\text{Lg}(10)=1$

4. ## Re: Logarithmic Explaination Needed

Originally Posted by Siron
$\log(x)=y \Leftrightarrow 10^{y}=x$

They use the logarithm with base 10 which is standard. If the base wasn't 10 the autors would have indicated that for sure. And it's not useful to write the base 10:
but then that formula is an equation you can equate the both sides and therefore become 10^y. in this question how come x=10^x?

5. ## Re: Logarithmic Explaination Needed

Do you understand $\log(x)=p \Leftrightarrow 10^p=x$ and $\log(y)=q \Leftrightarrow 10^q=y$ and so
$x\cdot y = 10^p\cdot 10^q=10^{p+q}$?

6. ## Re: Logarithmic Explaination Needed

Originally Posted by mushhall
but then that formula is an equation you can equate the both sides and therefore become 10^y. in this question how come x=10^x?
It's not and nobody said it was.
" $log(x)= p \leftrightarrow 10^p= x$" is two equations not one.
The $\leftrightarrow$ indicates "if one of those is true so is the other."

If $log(x)= p$ then $10^p= x$ and vice-versa. That is basically the definition of "logarithm".

7. ## Re: Logarithmic Explaination Needed

Thanks for the help.