Solve the following equations 710pt 42x316 2x2 810ptln xln 3

Solve the following equations 7(10pt). 4^2x+3=16 2^x-2 8(10pt).ln x=ln 3-ln(x-2) 7(10pt). log(x + 3) - 1 - >>oj(x - 2). 8(10pt). log(x-l)=log(1/3)+log(3x-5)

Solution

6. question is not visible in the Image

7. 4^(2x+3) = 16.2^(x-2)

LHS we can write : 4^(2x+3) = 2^2(2x+3)

RHS : 16.2^(x-2) = 2^4.2^(x-2) = 2^(x-2+4) = 2^(x+2)

So, we have  2^2(2x+3) = 2^(x+2)

We can equate the exponents if bases are same :

2(2x+3) = x+2

4x +6 = x+2

3x = -4

x = -4/3

8. lnx = ln3 - ln(x-2)

Use the log peroperty on RHS : lnA -lnB = ln(A/B)

lnx = ln( 3/x-2)

we equate the arguments on both sides i.e ln(A) = ln(B)---> A = B

So, x = 3/(x-2)

x^2 -2x -3 =0

x^2 -3x+x-3 =0

x(x-3) +1(x-3) =0

(x+1)(x-3) =0

x = -1 ;; x=3

we can neglect the -ve value as for lnx x>0

\'So, solution : x=3

7. log(x+3) = 1- log(x-2)

log(x+3) + log( x-2) = 1

Use the log property : logA +logB = log(A*B)

log(x+3)(x-2) =1

(x+3)(x-2) = 10^1

x^2 +x -6 =10

x^2 +x -16=0

use quadratic formula to solve for x:

x = ( -1 +/-sqrt( 1+64) )/2

= ( -1 +/- sqrt(65) )/2

neglecting -ve value of x as for logx x >0

x = (-1 +sqrt(65) )/2 = -0.5 +sqrt(65)