$$ \frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1) $$ $$ x=\text{?} $$

$$ -\frac{5}{4}\pm\frac{\sqrt{17}}{4} $$

$$ -\frac{5}{4}+\frac{\sqrt{17}}{4} $$

It is not possible to calculate

$$ -3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})= $$

$$ \ln(\frac{5}{4})^3+\log_5(\frac{7}{6})^3 $$

$$ \frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3= $$

$$ \log_411+\ln\frac{1}{3}\times\log_9 $$

$$ \log_{11}4+\log_3e\times\log6 $$

Power in logarithm

In order to solve a logarithm that appears in an exponent, you need to know all the logarithm rules including the sum of logarithms, product of logarithms, change of base rule, etc.

Solution steps:

Take the logarithm with the same base on both sides of the equation.
The base will be the original base - the one which the log power is applied to.
Use the rule
$log_a (a^x)=x$
Create a common base between the $2$ equation factors in order to determine the solution.
Solve the logs that can be solved and convert them to numbers.
Insert an auxiliary variable into the problem $T$ if needed
Go back to determine $X$ .

Detailed explanation

Start practice

Test yourself on power property of logorithms!

$2\log_38=$

Practice more now

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.

Let's begin by learning 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 take the form of an equation.

Solution methods:

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

Whilst an equation with a log in the exponent may appear confusing by following the previously mentioned steps we should be able to solve it easily.

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

Solution:

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 obtain the following:
$x^{1+log_2 4x}=log_x16$
A purely technical step - taking the log with the same base according to the original base on both sides of the equation.
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 as shown below.
$log_x (x^{1+log_2 4x} )=1+log_2 4x$
So that's what we'll do, insert the required data and continue the equation this way:
$1+log_2 4x=log_x 16$
Now we need to create a common factor that will help us to determine 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 obtain the following:
$log_x 16=\frac{log_2⁡16}{log_2⁡x }$
Insert the data into the equation once more as shown below.
$1+log_2 4x=\frac{log_2⁡16}{log_2⁡x }$
We continue with the 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 insert this into the equation and obtain:
$1+log_2 4+log_2 x=\frac{log_2⁡16}{log_2⁡x }$
Now we notice that some expressions can be converted to numbers. This way we will be left with a much less complex exercise that is entirely more manageable.
$log_2 4=2$
$log_2⁡16=4$
We proceed to substitute this into the equation and obtain the following:
$1+2+log_2 x=\frac{4}{log_2⁡x }$
$3+log_2 x=\frac{4}{log_2⁡x }$
Now we multiply by $log_2 x$ in order to remove the denominator from the equation and obtain:
$3\cdot log_2 x+(log_2 x)^2=4$
Now we'll use the helper factor $T$ and substitute it for the expression with the log.
As seen below: $log_2 x=t$
We substitute this into the equation and obtain:
$3t+t^2=4$
We move terms, factor the equation as follows:
$T_1=-4$
$T_2=1$
We insert the data in order to determine the $X$ and obtain:
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 a 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.

Join Over 30,000 Students Excelling in Math!

Endless Practice, Expert Guidance - Elevate Your Math Skills Today

Test your knowledge

Question 1

$3\log_76=$

Question 2

$\log_68=$

Question 3

$x\ln7=$

Examples with solutions for Power Property of Logorithms

Exercise #1

$3\log_76=$

Video Solution

Step-by-Step Solution

To simplify the expression $3\log_76$ , we apply the power property of logarithms, which states:

$a\log_b c = \log_b(c^a)$

Step 1: Identify the given expression: $3\log_76$ .

Step 2: Apply the power property of logarithms:

$3\log_76 = \log_7(6^3)$

Step 3: Calculate $6^3$ :

$6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$

Step 4: Substitute back into the logarithmic expression:

$\log_7(6^3) = \log_7216$

Therefore, the simplified expression is $\log_7216$ .

Comparing with the answer choices, the correct choice is:

$\log_7216$

Answer

$\log_7216$

Exercise #2

$2\log_38=$

Video Solution

Step-by-Step Solution

To solve this problem, let's simplify $2\log_3 8$ using logarithm rules.

Step 1: Recognize the expression form
The expression is of the form $a \cdot \log_b c$ , where $a = 2$ , $b = 3$ , and $c = 8$ .
Step 2: Apply the power property
According to the power property of logarithms, $2 \cdot \log_3 8$ can be simplified to $\log_3 (8^2)$ .
Perform the calculation
Calculate $8^2$ , which is $64$ .
Step 3: Simplify further
Therefore, we have $\log_3 64$ .

This is a straightforward application of the power property of logarithms. By applying this property correctly, we've simplified the original expression correctly.

Therefore, the simplified form of $2\log_3 8$ is $\log_3 64$ .

Answer

$\log_364$

Exercise #3

$\log_68=$

Video Solution

Step-by-Step Solution

To solve the problem $\log_6 8$ , we need to express the number 8 as a power of a base that simplifies the logarithm. We can write 8 as $2^3$ , because 8 equals 2 multiplied by itself three times.

Let's use the power property of logarithms, which is:

$\log_b (a^n) = n \log_b a$

Applying this property to $\log_6 8$ , we have:

$\log_6 8 = \log_6 (2^3)$

Using the power property, this becomes:

$\log_6 (2^3) = 3 \log_6 2$

Therefore, the expression for $\log_6 8$ in terms of $\log_6 2$ is:

$3 \log_6 2$ .

Answer

$3\log_62$

Exercise #4

$x\ln7=$

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow the steps outlined:

Step 1: Recognize that the expression $x \ln 7$ can be thought of in terms of the power property of logarithms, which helps reframe it into a single logarithm.
Step 2: Apply the formula $\ln(a^b) = b \ln a$ . This tells us that if we have something of the form $b \ln a$ , we can express it as $\ln(a^b)$ .
Step 3: Utilize the known expression and rule by substituting $a = 7$ and $b = x$ . Thus, $x \ln 7$ becomes $\ln(7^x)$ .

Therefore, the rewritten expression for $x \ln 7$ using logarithm rules is $\ln 7^x$ .

This matches choice 4 from the provided options.

Answer

$\ln7^x$

Exercise #5

$7\log_42<\log_4x$

Video Solution

Step-by-Step Solution

To solve the inequality $7\log_4 2 < \log_4 x$ , we will follow these steps:

Step 1: Simplify $7\log_4 2$ using the logarithm properties.
Step 2: Write the inequality $\log_4(2^7) < \log_4 x$ .
Step 3: Solve for $x$ by converting the logarithmic inequality into an exponential form.

Let's now proceed with these steps:

Step 1: Using the power property of logarithms, we have $7\log_4 2 = \log_4 (2^7)$ . This step simplifies the multiplication into a single logarithmic term.

Step 2: Using substitution in the inequality, we write it as $\log_4 (2^7) < \log_4 x$ .

Step 3: Since logarithms are one-to-one functions, we can conclude that if $\log_4 (2^7) < \log_4 x$ , then $2^7 < x$ . This results from the property $a^c = a^d \iff c = d$ where the bases are equal.

Therefore, the solution to the inequality is $\mathbf{2^7 < x}$ .

Answer

$2^7 < x$

Start practice

Power in logarithm

Power in logarithm

Test yourself on power property of logorithms!

Power in logarithm

Examples with solutions for Power Property of Logorithms

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Video Solution

Step-by-Step Solution

Answer

Exercise #3

Video Solution

Step-by-Step Solution

Answer

Exercise #4

Video Solution

Step-by-Step Solution

Answer

Exercise #5

Video Solution

Step-by-Step Solution

Answer

More Questions

Power Property of Logorithms