Solve the equation: log ( x + 2 ) x + log x ( x + 2 ) = 5 2

Given that log ( x + 2 ) x + log x ( x + 2 ) = 5 2 Now by applying the Interchange formula that is- log b a = 1 log a b , we get log ( x + 2 ) x + 1 log ( x + 2 ) x = 5 2 Now we assume that log ( x + 2 ) x = k Therefore, our equation becomes k + 1 k = 5 2 Multiply by 2 k , on both sides, we get ⇒ 2 k 2 + 2 = 5 k ⇒ 2 k 2 - 5 k + 2 = 0 Now we factorize the equation: 2 k 2 - 4 k - 1 k + 2 = 0 2 k ( k - 2 ) - 1 ( k - 2 ) = 0 ( k - 2 ) ( 2 k - 1 ) = 0 Therefore either k = 2 or k = 1 2 Now when we consider the value k = 2 , finding the solution is not possible as its discrimination is negative. Therefore, we consider the value k = 1 2 , of our solution. ⇒ log ( x + 2 ) x = 1 2 Now we apply the base change formula ⇒ log b a = c ⇒ b c = a We get ⇒ ( x + 2 ) 1 2 = x Now take square on both sides ⇒ x + 2 = x 2 ⇒ x 2 - x - 2 = 0 ⇒ x ( x - 2 ) + 1 ( x - 2 ) = 0 ⇒ x = 2 , - 1 As we know that the solution can not be less than zero Therefore x = 2 ,is the solution of log ( x + 2 ) x + log x ( x + 2 ) = 5 2

If log 5 ( 6 + 2 x ) + log 1 5 ( 1 + x 10 ) ≤ 1 then x lies in the interval

( − ∞ , 1 − 5 ) ∪ ( 1 + 5 ∞ )

E M B I B E

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.2

K. C. Sinha Solutions for Chapter: Appendix II : Logarithm, Exercise 2: EXERCISE 9.2

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Appendix II : Logarithm, Exercise 2: EXERCISE 9.2

Attempt the practice questions on Chapter 9: Appendix II : Logarithm, Exercise 2: EXERCISE 9.2 with hints and solutions to strengthen your understanding. TRIGONOMETRY FOR JEE MAIN AND ADVANCED solutions are prepared by Experienced Embibe Experts.

Questions from K. C. Sinha Solutions for Chapter: Appendix II : Logarithm, Exercise 2: EXERCISE 9.2 with Hints & Solutions

MEDIUM

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.1>Q 11

If $a, b, c$ be in G.P., show that $\log_{a} (x), \log_{b} (x), \log_{c} (x)$ are in H.P.

MEDIUM

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.2>Q 1

Solve the equation: $\log_{(x + 2)} x + \log_{x} (x + 2) = \frac{5}{2}$

MEDIUM

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.2>Q 5

Solve the equation: $\log_{10} (x^{2} - 12 x + 36) = 2$

EASY

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.2>Q 13

If $\log_{12} 18 = α$ and $\log_{24} 54 = β$ . Prove that : $α β + 5 (α - β) = 1$ .

HARD

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 9 - Appendix II : Logarithm>EXERCISE 9.2>Q 16

If $a, b, c$ be three such distinct positive numbers (none of them is $1$ ) that $(\log_{b} a \log_{c} a - \log_{a} a) + (\log_{a} b \log_{c} b - \log_{b} b) + (\log_{a} c \log_{b} c - \log_{c} c) = 0$ , then prove that $a b c = 1$ .