Question
Question: How do you evaluate \({\log _{\dfrac{1}{2}}}32\) ?...
How do you evaluate log2132 ?
Solution
First we write 32 in exponential form. 32 should be written in such a way that it is in the powers of 2. It is easier if we write it in powers of 2 so that we can cancel the logarithmic base with it. Use the laws of logarithms to remove the exponent and then simplify it.
Formulas:
Law of powers of logarithms, if we have the function, f(x)=loga(bc) . Then we can convert into power form as, f(x)=cloga(b) .
Law of the sum of logarithms, if we have a function, f(x)=logc(ab) . Then we can write it as f(x)=logca+logcb .
Law of base power of logarithms, if we have the expression, f(x)=log(bc)a . It can also be written as, f(x)=c1logba .
Complete step-by-step answer:
The given logarithmic expression is, log2132
Firstly we write 32 in exponential form, as the power of 2 .
Since 32=2×2×2×2×2
Now, on converting it into exponential form,
⇒32=25
Now put it back into a logarithmic expression.
⇒log21(25)
Now, consider the term.
By using the law of powers of logarithms,
If we have the function, f(x)=loga(bc) .Then we can convert into power form as, f(x)=cloga(b) .
Here a=21;b=2;c=5
On Substituting,
⇒5log21(2)
We can write the base of the logarithm which is 21 also as 2−1 for easy evaluation.
⇒5log(2−1)(2)
If we have the expression, f(x)=log(bc)a . It can also be written as, f(x)=c1logba .
Here, a=2;b=2;c=−1
On substituting,
⇒−15log22
If the base and the logarithm value is the same, they cancel out to get 1
⇒log22=1
⇒(−5)(1)
Now, putting it all together we get,
⇒−5
∴log2132=−5
Note:
One should ensure that the base of the given logarithms is the same before evaluating the expression using the laws of the logarithms. While using the subtraction law of logarithms ensure which term is written in the numerator and which in the denominator. Always the first term will be in the numerator whereas the second term in the denominator.