Test Index

Manipal Entrance Test 2013 Math Paper

© examsnet.com

Question : 68 of 80

Marks: +1, -0

\log_{2}(9-2^{x})=10^{\log(3-x)},

x

0
3
both (a) and (b)
0 and 6

Solution:

\log_{2}(9-2^{x})=10^{\log(3-x)}

⇒

\log_{2}(9-2^{x})=(3-x)

nbsp;

[ \because a^{\log_{a}b}=b ]

⇒

2^{3-x}=9-2^{x}

⇒

\frac{2^{3}}{2^{x}}=9-2^{x}

⇒

8=2^{x}\times(9-2^{x})

⇒

2^{2x}-2^{x}\times 9+8=0

Let

2^{x}=y,

then

y^{2}-9y+8=0

⇒

(y-8)(y-1)=0

⇒

y=8 \text{ or } y=1

⇒

2^{x}=2^{3} \text{ or } 2^{x}=2^{0}

⇒

x=3 \text{ or } x=0

But x = 3 does not satisfy the given equation, since log 0 is not defined.

© examsnet.com

Go to Question:

Prev Question

Next Question