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

##### 1 Answer

#### Explanation:

Use the change of base formula:

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

Here, we will use the arbitrary base

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

Recall that

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):}#