JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Application +4 | -1

Let

a=\log 25

and

b=\log 225

, then

\log \left(\frac{1}{81}\right)+\log \left(\frac{1}{2250}\right)

is equal to:

View answer and explanation

Correct option: C

2 a-3 b-1

Option C (

2a-3b-1

) is correct because by expressing the given logarithms in terms of a and b and simplifying, the expression

\log (\frac{{1}}{{81}})+\log (\frac{{1}}{{2250}})

simplifies to

2a-3b-1

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}