Question

solve this .......... que no. 22,23,24 ☺

Answer 1

Final Answer :
22 : x = 1, 4 and 0<= x <1
24 :x = 0 ,7/4 ,3/2 + √6

Steps and Understanding :

For Calculation see pic

Ans:22 (pic 1)
1) We will change base from 9 to 3 , by writing
9 = 3^2 and shift power to left side .
2) Then, we will just compare log inputs as bases are same.
3) We will take two cases,
Case : 1
x > = 1 .
Then, solve by opening with proper sign of modulus.
Case 2 :
0<x < 1
Then, same as above with proper sign of mod
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Ans:23 (bit conceptual Question) (pic 2,3)
But Don't worry,

Steps :
1) Open the logarithmic parts with product to sum rule and solve.
2) After, that multiply with (log (a, b) )^1/2 as given in Question.
3)Put
$log_{a}(b) = t$
Then, solve
After solving ,

4) We will reach at a stage in which we have to open mod by taking two cases.
Case 1 :
if t>=1
Case 2 :
if t< 1

Use this to get required expression,


Ans:24
Steps :
1) Let a=
$log(4 - x)$

b=
$log(x + \frac{1}{2} )$


2) We will get,
${a}^{2} + ab - 2 {b}^{2} = 0 \\ = > {a}^{2} + 2ab - ab - 2 {b}^{2} = 0 \\ = > (a + 2b)(a - b) = 0$
3) We will take both cases that is
a = b. and a + 2 b = 0

We will find x, y b such that x + 1/2 > 0
4-x>0

For Calculation see pic.