2log3x-log3(x-2)=2?

Question

When 3 is the base. I am pretty good at math but I cant solve this problem. Help would be appreciated, along with the work so I can understand how to do the problem.

Answer 1

2log3 x-log3 (x-2)=2

firstly, bring the 2 in front to become the power of x.
this rule is the power law rule
example: 3log2 x=log2 x^3

using this rule, u can easily work the qn out

using power law rule,
2log3 x=log3 x^2

lets continue..

log3 x^2-log3 (x-2)=2

since its of the same base, we can apply the quotient law

log3 (x^2/x-2)=2

can u identify which rule we need to use when we arrive to this step?

we must apply the change to index form rule.

log3 (x^2/x-2)=2

(x^2/x-2)=3^2
x^2/x-2=9
cross multiply it

9(x-2)=x^2
9x-18=x^2
bring it all to the left side to obtain a quadratic eqn.
x^2-9x+18=0
(x-6)(x-3)=0
therefore,
x=6 and x=3

here u go! the answer!

Answer 2

2 Log3

Answer 3

left side is equal to:
log3x^2-log3(x-2)=log3((x^2)/(x-2)
right side39
now you get x^2/(x-2)=9 multiply bith sides by x-2
x^2=9x-18
x^2-9x+18=0
(x-6)(x-3)=0
x=3 or x=6

properties used:
loga(b1/b2)=logab1 - logab2 where a is a base
logab^m=mlogab where a is a base

Answer 4

2 log x - log(x - 2) = 2 (all logs to base 3)
log x² - log (x - 2) = 2
log [x² / (x - 2)] = 2
x² / (x - 2) = 3²
x² = 9x - 18
x² - 9x + 18 = 0
(x - 6).(x - 3) = 0
x = 6, x = 3

Answer 5

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