CAT 2018 Slot 1 - QUANT Question 31

Question 31

If $$\log_{2}({5+\log_{3}{a}})=3$$ and $$\log_{5}({4a+12+\log_{2}{b}})=3$$, then a + b is equal to

Solution

$$\log_{2}({5+\log_{3}{a}})=3$$
=>$$5 + \log_{3}{a}$$ = 8
=>$$ \log_{3}{a}$$ = 3
or $$a$$ = 27

$$\log_{5}({4a+12+\log_{2}{b}})=3$$
=>$$4a+12+\log_{2}{b}$$ = 125
Putting $$a$$ = 27, we get
$$\log_{2}{b}$$ = 5
or, $$b$$ = 32

So, $$a + b$$ = 27 + 32 = 59
Hence, option A is the correct answer.


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