Question

How do you solve `log_3 (x) = 9log_x (3)`?

Answer 1

`x=1/27` or `27`

Explanation:

Use the change of base formula:

`log_a(b)=log_c(b)/log_c(a)`

Here, we will use the arbitrary base `c=10`, so we have

`log_a(b)=log_10(b)/log_10(a)=log(b)/log(a)`

Recall that `log_10(x)=log(x)`, as the common logarithm.

Applying the change of base formula, we see that

`log(x)/log(3)=(9log(3))/log(x)`

Cross multiply.

`log^2(x)=9log^2(3)`

Take the square root of both sides. Don't forget that this can be positive or negative!

`log(x)=+-3log(3)`

We can split this into the two equations:

`{:(log(x)=3log(3)," "" or "" ",log(x)=-3log(3)):}`

Rewrite the right hand side of both equations using the rule:

`b*log(a)=log(a^b)`

This gives us

`{:(log(x)=log(3^3)," "color(white)"or"" ",log(x)=log(3^-3)),(log(x)=log(27)," "color(white)"or"" ",log(x)=log(1/27)),(" "" "x=27," "color(white)"or"" "," "" "x=1/27):}`