Question

How do you simplify ${\log _2}({7^2}{.4^7})$?

Answer 1

Hint: Here we will use the property of logarithmic addition and the property of the power rule and then will simplify the given expression for the required resultant solution.

Complete step-by-step solution:
Take the given expression: ${\log _2}({7^2}{.4^7})$
Apply the property for the Product rule: ${\log _a}xy = {\log _a}x + {\log _a}y$
$ \Rightarrow {\log _2}({7^2}) + {\log _2}({4^7})$
Now, here Apply Power rule: ${\log _a}{x^n} = n{\log _a}x$ in the above expression –
$ \Rightarrow 2{\log _2}(7) + 7{\log _2}(4)$
Now, apply the square of the term where applicable in the above expression –
$ \Rightarrow 2{\log _2}(7) + 7{\log _2}{(2)^2}$
Apply Power rule: ${\log _a}{x^n} = n{\log _a}x$ in the above expression
$ \Rightarrow 2{\log _2}(7) + 7(2){\log _2}(2)$
By property, place ${\log _2}(2) = 1$in the above expression –
$ \Rightarrow 2{\log _2}(7) + 7(2)(1)$
Simplify the above expression –
$ \Rightarrow2{\log _2}(7) + 14$
This is the required solution.

Additional Information:
 Also refer to the below properties and rules of the logarithm.
Product rule: ${\log _a}xy = {\log _a}x + {\log _a}y$
Quotient rule: ${\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y$
Power rule: ${\log _a}{x^n} = n{\log _a}x$
 Base rule:${\log _a}a = 1$
Change of base rule: ${\log _a}M = \dfrac{{\log M}}{{\log N}}$

Thus the required is $2{\log _2}(7) + 14$.

Note: In other words, the logarithm is the power to which the number must be raised in order to get some other. Always remember the standard properties of the logarithm.... Product rule, quotient rule and the power rule. The basic logarithm properties are most important and the solution solely depends on it, so remember and understand its application properly. Be good in multiples and know the concepts of square and square root and apply accordingly.