Answer:
see below
Step-by-step explanation:
Question-6:
we are given a equation
[tex] \sf \displaystyle \: \log_{4}( - x) + \log_{4}( 6 - x) = 2[/tex]
to solve so
recall logarithm multiplication law:
[tex] \sf \displaystyle \: \log_{4}( - x \times (6 - x)) = 2[/tex]
simplify multiplication:
[tex] \sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = 2[/tex]
remember [tex]\displaystyle \log_{4}(4^2)=2[/tex]
so
[tex] \sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = \log_{4}( {4}^{2} ) [/tex]
cancel out [tex]\log_4[/tex] from both sides:
[tex] \sf \displaystyle \: - 6 x + {x}^{2} = {4}^{2} [/tex]
simplify squares:
[tex] \sf \displaystyle \: - 6 x + {x}^{2} = 16[/tex]
move left hand side expression to right hand side and change its sign:
since we are moving left hand side expression to right hand side there'll be only 0 left in the left hand side
[tex] \sf \displaystyle \: - 6 x + {x}^{2} - 16 = 0[/tex]
rewrite it to standard form i.e ax²+bx+c=0
[tex] \sf \displaystyle \: {x}^{2} - 6x - 16 = 0[/tex]
rewrite -6x as 2x-8x:
[tex] \sf \displaystyle \: {x}^{2} + 2x - 8x - 16 = 0[/tex]
factor out x and 8:
[tex] \sf \displaystyle \: x {(x}^{} + 2) - 8(x + 2) = 0[/tex]
group:
[tex] \sf \displaystyle \: (x - 8){(x}^{} + 2) = 0[/tex]
[tex] \displaystyle \: x = 8 \\ x = - 2 [/tex]
[tex] \therefore \: x = - 2[/tex]
Question-7:
move left hand side log to right hand side:
[tex] \displaystyle \: \log(x ) + \log(x - 21) = 2 [/tex]
use mutilation logarithm rule;
[tex] \displaystyle \: \log( {x}^{2} - 21x) = 2 [/tex]
[tex]\log(10^2)=2[/tex] so
[tex] \displaystyle \: \log( {x}^{2} - 21x) = \log({10}^{2} )[/tex]
cancel out log from both sides:
[tex] \displaystyle \: {x}^{2} - 21x = 100[/tex]
make it standard form:
[tex] \displaystyle \: {x}^{2} - 21x - 100= 0[/tex]
factor:
[tex] \displaystyle \: {(x} + 4)(x - 25)= 0[/tex]
so
[tex] \displaystyle \: x = 25[/tex]
