Question

I don't know how to do this question

Answer 1

a)
to solve a, the following rules are crucial:

(i)
`log_ab-log_ac=log_a (b)/(c)`

so the difference of 2 logarithms with the same base, is the logarithm of their division, preserving the same base.

(ii) if
`log_ab=log_ac` then b=c.

so if 2 logarithms with the same base are equal, then the arguments (b and c) are equal as well.

so

`log_3xy-log_3(x-1)=log_36 x^(2) -1`

apply rule (i):


`log_3 (xy)/(x-1)=log_3(6 x^(2) -1)`

apply rule (ii):


`(xy)/(x-1)=6 x^(2) -1`

now 'isolate' y:


`y=(6 x^(2) -1) ((x-1))/(x)`

b)
some more rules:

(iii)
`log_x a^(n)=nlog_x a`

(iv)
`log_a b= (1)/(log_b a)`

apply iii and iv:


`log_x25=log_x 5^(2)=2log_x 5=2 (1)/(log_5 x)=(2)/(log_5 x)`

then substitute
`log_5 x=u` in the equation:



`2u=5- (2)/(u)`


`2u=(5u-2)/(u)`


`2u^(2)=5u-2`


`2u^(2)-5u+2=0`

so now we have a quadratic equation of degree 2,

a=2, b=-5, c=2

the discriminant is
`b^(2)-4ac=25-16=9`, the root of it is 3

so the roots are:

`u_1= (-b+3)/(2a)= (5+3)/(2*2)= (8)/(4) =2`
and

`u_2= (-b-3)/(2a)= (5-3)/(2*2)= (2)/(4) = (1)/(2)`

finally, we convert u's to x's:


`log_5 x=u` means
`x=u^(2)`

so for u=2, x=4, and for u=1/2 we have x=1/4


Answers:

A) y=(6 x^{2} -1) \frac{(x-1)}{x}

B) solution set: {4, 1/4}