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Question

Question: How do you write \( {3^2} = 9 \) in logarithmic form?...

How do you write 32=9{3^2} = 9 in logarithmic form?

Explanation

Solution

Hint : In order to determine the value of the above question in logarithmic form ,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 this form and form your by=x{b^y} = x answer accordingly.

Complete step-by-step answer :
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Any logarithmic form logbx=y{\log _b}x = y when converted into equivalent exponential form results in
So in Our question we are given 32=9{3^2} = 9 and if compare this with logbx=y{\log _b}x = y we get
b=3 y=2 x=9   b = 3 \\\ y = 2 \\\ x = 9 \;
Hence the logarithmic form of 32=9{3^2} = 9 will be equivalent to log39=2{\log _3}9 = 2 .
Therefore, our desired answer is log39=2{\log _3}9 = 2 .
So, the correct answer is “ log39=2{\log _3}9 = 2 ”.

Note : 1. Value of the 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}(\dfrac{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}