Question

How do you solve $\log x + \log 8 = 2$?

Answer 1

Consider $\log a + \log b = \log ab$, here we have to find the value of a i.e., $x$ as the value of b and ab is given.

Formula used:
$\log a + \log b = \log ab$

Complete step by step answer:
To solve $\log x + \log 8 = 2$, let us take $\log a + \log b = \log ab$ into consideration and solve the equation.
$\log x + \log 8 = 2$
$\log 8x = 2$
Here log implies ${\log _{10}}$
${\log _{10}}8x = 2$
$8x = {10^2}$
$8x = 100$
Therefore, the value of x is
$x = \dfrac{{100}}{8}$
$x = \dfrac{{25}}{8}$
In decimal the value of x is
$x = 12.5$

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.
Logarithms to the base ‘e’ are called natural logarithms. The constant e is approximated as 2.7183.
Natural logarithms are expressed as ln x which is the same as log e
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: The logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication.