Question

Can SOMEBODYYY please do the maths!!! PLeassssssssssssseeeeeeee

Answer 1

Answer:

see below

Step-by-step explanation:

Question-6:

we are given a equation

$\sf \displaystyle \: \log_{4}( - x) + \log_{4}( 6 - x) = 2$

to solve so

recall logarithm multiplication law:

$\sf \displaystyle \: \log_{4}( - x \times (6 - x)) = 2$

simplify multiplication:

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = 2$

remember $\displaystyle \log_{4}(4^2)=2$

so

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = \log_{4}( {4}^{2} )$

cancel out $\log_4$ from both sides:

$\sf \displaystyle \: - 6 x + {x}^{2} = {4}^{2}$

simplify squares:

$\sf \displaystyle \: - 6 x + {x}^{2} = 16$

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

$\sf \displaystyle \: - 6 x + {x}^{2} - 16 = 0$

rewrite it to standard form i.e ax²+bx+c=0

$\sf \displaystyle \: {x}^{2} - 6x - 16 = 0$

rewrite -6x as 2x-8x:

$\sf \displaystyle \: {x}^{2} + 2x - 8x - 16 = 0$

factor out x and 8:

$\sf \displaystyle \: x {(x}^{} + 2) - 8(x + 2) = 0$

group:

$\sf \displaystyle \: (x - 8){(x}^{} + 2) = 0$

$\displaystyle \: x = 8 \\ x = - 2$

$\therefore \: x = - 2$

Question-7:

move left hand side log to right hand side:

$\displaystyle \: \log(x ) + \log(x - 21) = 2$

use mutilation logarithm rule;

$\displaystyle \: \log( {x}^{2} - 21x) = 2$

$\log(10^2)=2$ so

$\displaystyle \: \log( {x}^{2} - 21x) = \log({10}^{2} )$

cancel out log from both sides:

$\displaystyle \: {x}^{2} - 21x = 100$

make it standard form:

$\displaystyle \: {x}^{2} - 21x - 100= 0$

factor:

$\displaystyle \: {(x} + 4)(x - 25)= 0$

so

$\displaystyle \: x = 25$