Question

Solve for x:

Log base 3 (x)^2 - log base 3 (x)^4 + 3 = 0

Answer 1

Am I supposed to let a = log base 3 (x)^2 ???

Answer 2

So that it will become -a + a +3 = 0???

Answer 3

\[\Large\bf\sf \color{#DD4747}{\log_3 x^2- \log_3 x^4}+3\quad=\quad 0\]We can combine these first two logs using a rule of logarithms.

Answer 4

Log base 3 (x^2/x^4)

Answer 5

Ok good, now we'll want to deal with this 3, it's a little tricky.
\[\Large\bf\sf \log_3\frac{x^2}{x^4}+\color{#DD4747}{3}\quad=\quad 0\]

Answer 6

Log base 3 (27)???

Answer 7

Mm yes very good!\[\Large\bf\sf \color{#DD4747}{\log_327=3}\]

Answer 8

That's how I converted it

Answer 9

\[\Large\bf\sf \log_3\frac{x^2}{x^4}+\log_327=0\]

Answer 10

Remember the rule for adding logs?

Answer 11

Multiplication

Answer 12

Ill see what answer ill get

Answer 13

What do we do with the zero

Answer 14

\[\Large\bf\sf \log_3\frac{27x^2}{x^4}\quad=\quad 0\]You can either do the thing you did earlier, the sneaky tricky, turning it into a log of base 3.
Or you can convert to exponential form from here if you remember how.

Answer 15

Could you please show me

Answer 16

\[\Large\bf\sf \log_31=0\]\[\Large\bf\sf \log_3 3=1\]\[\Large\bf\sf \log_39=2\]\[\Large\bf\sf \log_327=3\]You converted 3 earlier using this last one, right?
See how we can use the same trick for 0?

Answer 17

Now I see. Yes I remember

Answer 18

And how can we doit using exponential