We will use the following property of logarithms (logarithm of a power)

`log_a b^n=n log_a b`
**(1)**

**1.**

`log_3 x^3=(log_3 x)^2`

Now by using (1) on the left hand side we get

`3log_3x=(log_3 x)^2`

Now we make substitution `t=log_3x`

`3t=t^2`

`t^2-3t=0`

`t(t-3)=0`

From the above line we have 2 solutions `t_1=0` and `t_2=3`. Now we return to our substitution by putting `t_1` and `t_2` instead of `t`.

`0=log_3x_1`

`3^0=x_1`

`x_1=1` **<-- First solution**

`3=log_3 x_2`

`3^3=x_2`

`x_2=27` **<-- Second solution**

**2.**

This is very similar to previous equation

`log_2 x^4=(log_2x)^2`

Again we use (1) to get

`4log_2x=(log_2x)^2`

Substitution `t=log_2x`

`4t=t^2`

`t^2-4t=0`

`t(t-4)=0=> t_1=0,\ t_2=4 `

`0=log_2x_1`

`2^0=x_1=>x_1=1` **<--First solution**

`4=log_2x_2`

`2^4=x_2=> x_2=16` **<--Second solution**

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