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Question

Question: How is this \( \log x = 4 \) done?...

How is this logx=4\log x = 4 done?

Explanation

Solution

Hint : In order to determine the value of the above question, we will convert the expression into exponential form, and to do so use the definition of logarithm that the logarithm of the form logbx=y{\log _b}x = y is when converted into exponential form is equivalent to by=x{b^y} = x ,so compare with the given logarithm value with this form ,we will get your required answer.

Complete step-by-step answer :
To solve the given logarithmic function logx=4\log x = 4 , we must know the properties of logarithms and with the help of them we are going to rewrite our question.
If there is no base given, it means it is a logarithm of base 1010 .
log10x=4{\log _{10}}x = 4
Let’s convert this into its exponential form to remove log10{\log _{10}} .
Any logarithmic form logbX=y{\log _b}X = y when converted into equivalent exponential form results in by=X{b^y} = X
So in Our question we are given log10x=4{\log _{10}}x = 4 and if compare this with logbx=y{\log _b}x = y we get
b=10 y=4 X=x   b = 10 \\\ y = 4 \\\ X = x \;
log10x=4 x=104 x=10000   \Rightarrow {\log _{10}}x = 4 \\\ \Rightarrow x = {10^4} \\\ \Rightarrow x = 10000 \;
Therefore, solution to equation logx=4\log x = 4 is equal to x=10000=104x = 10000 = {10^4}
So, the correct answer is “x=10000=104x = 10000 = {10^4}”.

Note : 1. Value of constant” e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
logb(mn)=logb(m)+logb(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
logb(mn)=logb(m)logb(n){\log _b}(\frac{m}{n}) = {\log _b}(m) - {\log _b}(n)
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlogm=logmnn\log m = \log {m^n}