Get 2 FREE Instant-Explanations on Filo with code FILOAPP

If a>1 is a real number, then the number of roots of the equat

Question

If $a > 1$ is a real number, then the number of roots of the equation $a^{2 l o g_{2} x}$ = 5+4 $x^{l o g_{2} a}$ has

infinite

Views: 989 students

Found 8 tutors discussing this question

Levi Discussed

a > 1

is a real number, then the number of roots of the equation

a^{2 l o g_{2} x}

= 5+4

x^{l o g_{2} a}

has

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

We have,

a^{2 l o g_{2} x} = 5 + 4 x^{l o g_{2} a}

\Rightarrow a^{2 l o g_{2} x} = 5 + 4 a^{l o g_{2} x}, [∵ a^{l o g_{b} c} = c^{l o g_{b} a}]

(x^{l o g_{2} a})^{2} = 5 + 4 x^{l o g_{2} a}

Now, let

p = x^{l o g_{2} a}

, giving us,

p^{2} = 5 + 4 p

(p - 5) (p + 1) = 0

p = 5 or - 1

p = x^{l o g_{2} a} = a^{l o g_{2} x}

So,

p > 0

, because

a > 1

, so power of positive number is always positive.

So,

p \neq = - 1

∴ p = 5

\Rightarrow x^{l o g_{2} a} = 5 \Rightarrow lo g_{2} a . lo g_{2} x = lo g_{2} 5 \Rightarrow lo g_{2} x = \frac{lo g _{2} 5}{lo g _{2} a} = \frac{lo g 5}{lo g 2} \times \frac{lo g 2}{lo g a} = \frac{lo g 5}{lo g a} = lo g_{a} 5 \Rightarrow x = 2^{l o g_{a} 5}

So, Number of roots is

1

So, option

D

is correct

Was this solution helpful?

150

Report

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our maths tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	If $a > 1$ is a real number, then the number of roots of the equation $a^{2 l o g_{2} x}$ = 5+4 $x^{l o g_{2} a}$ has
Answer Type	Text solution:1
Upvotes	150