Question

How do you solve for $x$ and $y$ if $\log x-\log =2$ and $\log x+\log y=0$

Answer 1

Hint: We have to find $n$ and $y$ intercepts so we can use different logarithmic law. Here we can use logarithmic Quotient rule ${{\log }_{b}}\left( x \right)-{{\log }_{b}}\left( 4 \right)={{\log }_{b}}\left( \dfrac{n}{4} \right)$
Where, $b$ is the base here we will assume log base $10$ because here there is no given base. So that’s the reason we will use base $10.$
First isolate the $\log x$ from any one equation then this value to another equation..

Complete step-by-step answer:
Given $\log x-\log y=2$ as per question,
$\Rightarrow$ $\log x-\log y=2...(i)$
$\Rightarrow$ $\log x+\log y=0...(ii)$
Isolate the $\log x$ in the equation $(i)$
$\log x=2+\log y$ we will name this equation $(iii)$ substitute the value of $\log x$ in the equation $(ii)$
$\Rightarrow$ $\log x+\log y=0$
Put $\log x=2+\log y$
Now, $2+\log y+\log y=0$
Then will become
$\Rightarrow$ $2+2\log y=0$
We can write $2+2\left( \log y \right)=0$ as
$\Rightarrow$ $2\left( 1+\log y \right)=0$
$\Rightarrow$ $2\left( 1+\log y \right)=0$
Now, $\log y=-1$
$\Rightarrow$ $y={{10}^{-1}}$
We can write $y={{10}^{-1}}$ as $y=\dfrac{1}{10}$
Therefore $y=\dfrac{1}{10}$
Substitute the value of $y$in equation $(iii)$
$\Rightarrow$ $\log x\pm 2+\log y$
$\Rightarrow$ $y=\dfrac{1}{10}$
We will get $\log x=2\log \left( \dfrac{1}{10} \right)$
Take the log equation on one side.
$\Rightarrow$ $\log x-\log \left( \dfrac{1}{10} \right)=2$
We input logarithm quotient rule ${{\log }_{b}}\left( x \right)-{{\log }_{b}}\left( y \right)$
$\Rightarrow$ ${{\log }_{b}}\left( \dfrac{x}{4} \right)$
So, will apply logarithm quotient rule we will get,
\[\log \left( \dfrac{x}{\dfrac{1}{10}} \right)=2\]
$\Rightarrow$ $10x={{10}^{2}}$
$\Rightarrow$ $10x=100$
$\Rightarrow$ $x=10$
So, the solution of $\left( x,4 \right)$is $\left( 10,\dfrac{1}{10} \right)$

We assumed the logarithms base as $10.$

Additional Information:
The logarithm with the base $10$ is called the common logarithm. To get the logarithm of a number $n$ find the number $x$ that when the base is raised to that power the resulting value is $n$ for this problem we have,
$\Rightarrow$ ${{\log }_{10}}10=x$
$\Rightarrow$ ${{10}^{x}}=10$
$\Rightarrow$ ${{10}^{x}}={{10}^{1}}$
$\Rightarrow$ $x=1$
Therefore the common logarithm of $10$ is $1$ by condensing the log expression on the left into a singer logarithm using the product rule. What we want is to have a single log expression on each side of the question be ready though to solve for a quadratic since $x$ Will have a power of $2.$ Solve the quadratic equation using the factoring method. There are two main reasons to use logarithmic scales in charts and graphs. The first is to respond to Skegness towards the large values. i.e. in which one a few points are much larger than the bulk of the data the second is to show present change on multiplication factors.

Note:
It is okay to have values of $x$ such as positive, $0$ and negative numbers however it is not allowed to have a logarithm of a negative number or a logarithm of zero. When substituted or evaluated into the original logarithm equation. Solve the problem carefully.