Question

How do you solve \[\log x = {x^2} - 2\]?

Answer 1

Hint:In the given equation we need to find the value of x, as the equation is \[\log x = {x^2} - 2\], applying logarithmic properties we can solve this equation.

Complete step by step answer:
Let us write the given equation as
\[\log x = {x^2} - 2\]
Applying logarithmic properties, let us raise 10 to the power of each side of the equation given both in LHS and RHS part
\[{10^{\log x}} = {10^{{x^2} - 2}}\]
Hence to find the value of x we have
\[x = {10^{{x^2} - 2}}\]
As in the equation no logarithmic value is given to the terms.
Therefore, we get the value of x as \[x= {10^{{x^2} - 2}}\].
Additional information:
Rules of Logarithms
Logarithms are a very disciplined field of mathematics. They are always applied under certain rules and regulations.
Given that \[{a^n} = b \Leftrightarrow {\log _a}b = n\], the logarithm of the number b is only
defined for positive real numbers \[ \Rightarrow a > 0\left( {a \ne 1} \right),{a^n} > 0\]
The logarithm of a positive real number can be negative, zero or positive.
Logarithmic values of a given number are different for different bases.
Logarithms to the base a 10 are referred to as common logarithms. When a logarithm is written without a subscript base, we assume the base to be 10.
The logarithmic value of a negative number is imaginary and the logarithm of any positive number to the same base is equal to 1.
\[{a^1} = a \Rightarrow {\log _a}a = 1\]
The logarithm of 1 to any finite non-zero base is zero.
\[{a^0} = 1 \Rightarrow {\log _a}1 = 0\]

Note: Here in the equation to find the value x, as there is no exponential term given in the equation and based on the logarithmic function, we can apply the respective formula and solve any kind of equation provided.