Question

How do you solve \[{\log _2}\left( {x + 1} \right) = 0\]?

Answer 1

Hint: To solve \[{\log _2}\left( {x + 1} \right) = 0\], we will use the definition of \[\log \] function i.e., if \[{\log _b}a = c\] then \[{b^c} = a\] where \[a,b > 0\] and \[b \ne 1\]. Using this we will write \[{\log _2}\left( {x + 1} \right) = 0\] as \[\left( {{2^0} = x + 1} \right)\]. Then using the property of exponents \[{a^0} = 1\] we will simplify it to find the result.

Complete step by step answer:
We have to solve \[{\log _2}\left( {x + 1} \right) = 0\]. Here the base of \[\log \] is \[2\].
As we know, by the definition of \[\log \] function that if \[{\log _b}a = c\] then \[{b^c} = a\] where \[a,b > 0\] and \[b \ne 1\].
Using this we can write \[{\log _2}\left( {x + 1} \right) = 0\] as
\[ \Rightarrow {2^0} = x + 1\]
Now from the property of exponents we know that zero raised to any number is equal to one. So, we get
\[ \Rightarrow 1 = x + 1\]
On rewriting the equation, we get
\[ \Rightarrow x + 1 = 1\]
Subtracting \[1\] from both the sides, we get
\[ \Rightarrow x = 0\]
Therefore, on solving \[{\log _2}\left( {x + 1} \right) = 0\], we get \[x = 0\].

Note:
It is very important to note that \[\log x\] denotes that the base is \[10\] and \[\ln x\] denotes that base is \[e\]. Also, \[{\log _b}x\] is only defined when \[b\] and \[x\] are two positive real numbers and \[b\] is not equal to \[1\]. Logarithm base \[10\] i.e., \[b = 10\] is called decimal or common logarithm, logarithm base \[e\] is called natural logarithm and binary logarithm uses base \[2\].
Properties of Logarithmic functions are as follows:
\[(1)\] The Product Rule: \[{\log _b}(mn) = {\log _b}(m) + {\log _b}(n)\]
\[(2)\] The Quotient Rule: \[{\log _b}\dfrac{m}{n} = {\log _b}m - {\log _b}n\]
\[(3)\] The Power Rule: \[{\log _b}\left( {{m^p}} \right) = p{\log _b}m\]
\[(4)\] The Zero Exponent Rule: \[{\log _b}1 = 0\]
\[(5)\] Change of Base Rule: \[{\log _b}x = \dfrac{{{{\log }_a}x}}{{{{\log }_a}b}}\]