Question
Question: How do you solve \[\log x - \log 2 = 1\]...
How do you solve logx−log2=1
Solution
When one term is raised to the power of another term, the function is called an exponential function, for example a=xy . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example is y=logxa that is a logarithm function. These functions also obey certain rules called laws of the logarithm, using these laws we can write the function in a variety of ways. Using the laws of the logarithm, we can solve the given equation and find out the value of x.
Complete step by step answer:
We are given that logx−log2=1
We know that
logx−logy=logyx ⇒logx−log2=log2x
Using this in the given equation, we get –
log2x=1
We also know that –
if,lognx=a ⇒x=na
So,
$
{\log _{10}}\dfrac{x}{2} = 1 \\
\Rightarrow \dfrac{x}{2} = {(10)^1} \\
\Rightarrow x = 10 \times 2 \\
\Rightarrow x = 20 \\
$
Hence, if logx−log2=1 then x=20 .
Note: The important condition while applying the laws of the logarithm is that the base of the logarithm functions involved should be the same in all the calculations. When we are given a function without any base like logx then we take the standard base that is 10, so the base of both logx and log2 is 10, as the base of both these functions is the same, we can apply the logarithm laws in the given question. There are three laws of the logarithm, one of addition, one of subtraction and the other to convert logarithm functions to exponential functions. In the given question, we have used the law for subtraction as the two logarithm functions are in subtraction with each other then we applied the third law to find the value of x.