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# How do you solve $\log \left( x \right)+\log \left( 2x \right)=10?$

Last updated date: 20th Jul 2024
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Hint: Here in the given logarithms we have to use the product rule. The product rule says that the logarithm of a product is the sum of logs of its factor. The product rule is ${{\log }_{b}}\left( MN \right)={{\log }_{b}}\left( M \right)+{{\log }_{b}}\left( N \right).$ Use this rule for rewriting the logarithmic expression when we shove any expression using product rule bases of the given logarithm must be the same.

Complete step by step solution:
Here we have,
Given: $\log \left( x \right)+\log \left( 2x \right)=10$
By using rule,
$\log m+\log n=\log mn$
Here, $m=x$ and $n=2x$
Multiply $m,n$ with each other we get,
$\therefore \log \left( x.2x \right)=10$
$\therefore \log \left( 2{{x}^{2}} \right)=10$
$\therefore 2{{x}^{2}}={{10}^{10}}$
${{x}^{2}}=\dfrac{{{10}^{10}}}{2}$
$x=\sqrt{\dfrac{{{10}^{10}}}{2}}$
$\therefore x=\pm \dfrac{{{10}^{5}}}{\sqrt{2}}$
$\therefore x=50000\sqrt{2},-500000\sqrt{2}$

Hence,By solving the $\log \left( x \right)+\log \left( 2x \right)=10$, We get the values, $50000\sqrt{2}$ and $-50000\sqrt{2}$

In any exponential function and logarithm any number can be the base. However there are two bases used and also there are special names for this logarithm. In the scientific calculator you have the keys for finding the log and ln easily. Sometimes the input for the logarithm exponent on the base is more complicated. Than just a single variable. As an example we take $\log \left( 3x \right)$ Where we have $x=4$ the answer get $1.079.$ but if you didn’t use the brackets that is $\log 3x.$ then the calculator find $\log 3$ and multiply it to $4.$ Then we get $1.908.$ Which is the incorrect answer.
Note: We have the option in the calculator so use the calculator to determine the logarithm or power base $'e'$ Check out the given logarithm that which identity or rule suits to it and then apply it in the given logarithm. Simplify it but you get both the value as positive and negative values. Write both as an answer. We have the definition of a logarithm that is ${{\log }_{b}}\left( a \right)=C={{b}^{c}}=a$ Where as $b$ is base and $c$ is exponent and $a$ is argument. If you have any exponential form by using the definition of $\log$ you can easily simplify it in logarithmic form. These are the important tips you have to remember before solving this type of problem.