Question

I don't know how to do this question

Answer 1

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

(i) $log_ab-log_ac=log_a \frac{b}{c}$

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):

$log_3 \frac{xy}{x-1}=log_3(6 x^{2} -1)$

apply rule (ii):

$\frac{xy}{x-1}=6 x^{2} -1$

now 'isolate' y:

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

b) 
some more rules:

(iii) $log_x a^{n}=nlog_x a$

(iv) $log_a b= \frac{1}{log_b a}$

apply iii and iv:

$log_x25=log_x 5^{2}=2log_x 5=2 \frac{1}{log_5 x}=\frac{2}{log_5 x}$

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


$2u=5- \frac{2}{u}$

$2u=\frac{5u-2}{u}$

$2u^{2}=5u-2$

$2u^{2}-5u+2=0$

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:
$u_1= \frac{-b+3}{2a}= \frac{5+3}{2*2}= \frac{8}{4} =2$
and
$u_2= \frac{-b-3}{2a}= \frac{5-3}{2*2}= \frac{2}{4} = \frac{1}{2}$

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}