Power in logarithm

Very important - review all logarithm rules - starting from the definition of log, multiplication, addition and change of log base. This topic includes all subjects within it.

First, let's learn the following rule:
$log_a (a^x)=x$
This rule will help us eliminate long and cumbersome expressions later on, so remember it.
Exercises where the logarithm appears in an exponent usually come as an equation.

Solution methods:

1. Take the logarithm with the same base on both sides of the equation.
   The base will be the original base - the one on which the log power is applied.
   This is a completely technical step where we write log on both sides of the equation. The content of the log will be the original data.
2. Use the rule $log_a (a^x)=x$
3. Create a common base between the $2$ equation factors to reach a solution. If one of the bases is X, we will want to convert it to the other base.
4. Solve the logs that can be solved and convert them to numbers.
5. Substitute an auxiliary variable $T$ if needed
6. Return to find $X$.

We know, it looks a bit confusing and cumbersome, but let's see how while solving we'll follow the steps and easily solve an equation with a log in the exponent.

Here is the exercise:
$x^{1+log_2 4x}=16$

Solution:

1. In the first step, we take the logarithm with the same base on both sides of the equation.
   The base is the original base = in this exercise $X$ (on which the $log$ power is applied)
   We get:
   $x^{1+log_2 4x}=log_x16$
   Simply technical work - taking log with the same base according to the original base on both sides of the equation.
2. Now let's remember the important rule we learned at the beginning of the article:
   $log_a (a^x)=x$
   According to the rule, we can remove the entire expression on the left side and leave only the exponent. Because according to the rule:
   $log_x (x^{1+log_2 4x} )=1+log_2 4x$
   So that's what we'll do, substitute and continue the equation this way:
   $1+log_2 4x=log_x 16$
3. Now we need to create a common factor that will lead us to the solution.
   How will we do that?
   Let's remember the law of changing the logarithm base:
   $log_aX=\frac{log_{the~base~we~want~to~change~to}X}{log_{the~base~we~want~to~change~to}a}$
   We want to convert the logarithm in base $X$ to base $2$. Using the rule we get:
   $log_x 16=\frac{log_2⁡16}{log_2⁡x }$
   Now we substitute in the equation and get:
   $1+log_2 4x=\frac{log_2⁡16}{log_2⁡x }$
   We continue with logarithm laws -
   According to the multiplication law:
   $log_a⁡(x\cdot y)=log_a⁡x+log_a⁡y$
   Therefore on the left side we'll change
   $1+log_2 4x=$
   to:
   $1+log_2 4+log_2 x$
   We substitute in the equation and get:
   $1+log_2 4+log_2 x=\frac{log_2⁡16}{log_2⁡x }$
4. Now we notice that some expressions can be converted to numbers. This way we can get a much less complex exercise that's much more pleasing to the eye.
   $log_2 4=2$
   $log_2⁡16=4$
   Now we substitute this in the equation and get:
   $1+2+log_2 x=\frac{4}{log_2⁡x }$
   $3+log_2 x=\frac{4}{log_2⁡x }$
   Now we multiply by $log_2 x$ to remove the denominator from the equation and get:
   $3\cdot log_2 x+(log_2 x)^2=4$
5. Now we'll use the helper factor $T$ and substitute it for the expression with the log.
   We get that: $log_2 x=t$
   We substitute in the equation and get:
   $3t+t^2=4$
   We move terms, factor the equation and get that:
   $T_1=-4$
   $T_2=1$
   We substitute to find $X$ and get:
   First solution when $T_1=-4$
   $log_2 x=-4$
   $​​​​​​​x=2^{-4}=0.0625$
   Second solution when $T_2=1$
   $log_2 x=1$
   $x=2^1=2$

Note - You may encounter exercises without base $X$ but with a numerical base instead. These are usually simpler and easier exercises that don't require the auxiliary variable $T$. However, if you know how to solve exponential logarithm exercises with base $X$, solving with a regular base will certainly be easier. The solution method is identical.