Solveeit Logo

Question

Question: How do you express as a single logarithm and simplify \( \left( {\dfrac{1}{2}} \right){\log _a}.x...

How do you express as a single logarithm and simplify
(12)loga.x+4loga.y3loga.x\left( {\dfrac{1}{2}} \right){\log _a}.x + 4{\log _a}.y - 3{\log _a}.x ?

Explanation

Solution

Hint : Here we will use the logarithmic properties to simplify the given expression. We will use Power rule, quotient rule and the product rule. Then will simplify for the required resultant value.

Complete step-by-step answer :
Take the given expression:
(12)loga.x+4loga.y3loga.x\left( {\dfrac{1}{2}} \right){\log _a}.x + 4{\log _a}.y - 3{\log _a}.x
Using power rule in the above expression: logaxn=nlogax{\log _a}{x^n} = n{\log _a}x
=loga.x12+loga.y4loga.x3= {\log _a}.{x^{\dfrac{1}{2}}} + {\log _a}.{y^4} - {\log _a}.{x^3}
Now using the Product rule: logaxy=logax+logay{\log _a}xy = {\log _a}x + {\log _a}y which states that bases are same and then is plus sign then it is multiplied.
=loga.x12.y4loga.x3= {\log _a}.{x^{\dfrac{1}{2}}}.{y^4} - {\log _a}.{x^3}
Again, using the law of Quotient rule: logaxy=logaxlogay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y which states that when bases are same and there is subtraction sign in between then it is applied as the division.
=loga.x12.y4x3= {\log _a}.\dfrac{{{x^{\dfrac{1}{2}}}.{y^4}}}{{{x^3}}}
Now, using the law of power and exponent and simplify for the power of “x”. when power is moved from numerator to denominator, sign also changes. Positive power becomes negative and vice-versa.
=loga.y4x312= {\log _a}.\dfrac{{{y^4}}}{{{x^{3 - \dfrac{1}{2}}}}}
Simplify taking LCM (Least common multiple) for the denominator of the above expression.
=loga.y4x52= {\log _a}.\dfrac{{{y^4}}}{{{x^{\dfrac{5}{2}}}}}
Hence, (12)loga.x+4loga.y3loga.x=loga.y4x52\left( {\dfrac{1}{2}} \right){\log _a}.x + 4{\log _a}.y - 3{\log _a}.x = {\log _a}.\dfrac{{{y^4}}}{{{x^{\dfrac{5}{2}}}}}
This is the required solution.
So, the correct answer is “loga.y4x52{\log _a}.\dfrac{{{y^4}}}{{{x^{\dfrac{5}{2}}}}} ”.

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.
Also refer to the below properties and rules of the logarithm.
Product rule: logaxy=logax+logay{\log _a}xy = {\log _a}x + {\log _a}y
Quotient rule: logaxy=logaxlogay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y
Power rule: logaxn=nlogax{\log _a}{x^n} = n{\log _a}x
Base rule: logaa=1{\log _a}a = 1
Change of base rule: logaM=logMlogN{\log _a}M = \dfrac{{\log M}}{{\log N}}