Question

Which of the following is not true?
A. \[{\log _2}8 = 3\]
B. \[{\log _2}1 = 0\]
C. \[{\log _2}2 = 1\]
D. \[{\log _2}\dfrac{1}{8} = - 3\]
E. \[{\log _2}\left( { - 1} \right) = \dfrac{1}{2}\]

Answer 1

Hint: Logarithm is the inverse function of exponentiation. It means that the logarithm of a given number \[y\] is the exponent to which another number, the base \[b\], must be raised to produce that number \[y\]. To solve this question, we will use the definition of logarithm and check all the options one by one.

Formula used:
If \[{\log _b}x = y\], then \[x = {b^y}\]

Complete step by step answer:
We have the first option as;
\[{\log _2}8 = 3\]
It is in the form of \[{\log _b}x = y\], so, we will check whether \[x = {b^y}\], or not. We have;
\[x = 8,b = 2,y = 3\]
Now,
\[ \Rightarrow {b^y} = {2^3}\]
\[ \Rightarrow {b^y} = 8 = x\]
Hence, the first option is true.

Second option: \[{\log _2}1 = 0\]
Using the formula as above. We have;
\[{2^0} = 1\], Which is the same value as the argument.
Hence, the second option is correct.

Third option: \[{\log _2}2 = 1\]
When we put \[{2^1}\], we get the value \[2\]. Which is the same value as the argument.
Hence, the third option is true.

Fourth option: \[{\log _2}\dfrac{1}{8} = - 3\]
When we put \[{2^{ - 3}}\], we get the value \[\dfrac{1}{8}\]. Which is the same value as the argument.
Hence this option is also true.

Fifth option: \[{\log _2}\left( { - 1} \right) = \dfrac{1}{2}\]
When we put \[{2^{\dfrac{1}{2}}}\], the value we get is \[1.414\]. But the value in the argument is \[ - 1\]. So, this option is not true.

Therefore,option A is true.

Note: We should note that argument and base should be greater than zero and base should not be equal to one. Mathematically,
\[x > 0,b > 0,b \ne 1\].
So, now if we observe the options carefully, we can see that in the last option this rule is not followed. There the base is negative. So, we can directly tell that the last option is not true.