Power Property of Logarithms Practice Problems & Solutions

Master logarithm power rules with step-by-step practice problems. Learn to solve equations with logs in exponents using proven techniques and examples.

📚Master Logarithm Power Property Problems with Expert Guidance
  • Apply the fundamental rule log_a(a^x) = x to simplify complex expressions
  • Solve equations with logarithms appearing in exponents step-by-step
  • Use change of base formula to create common factors in logarithmic equations
  • Convert logarithmic expressions to numerical values using power properties
  • Substitute auxiliary variables to solve quadratic logarithmic equations
  • Verify solutions and identify extraneous answers in logarithmic problems

Understanding Power Property of Logorithms

Complete explanation with examples

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:

  1. 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.
  2. Use the rule
    loga(ax)=xlog_a (a^x)=x
  3. Create a common base between the 22 equation factors in order to determine the solution.
  4. Solve the logs that can be solved and convert them to numbers.
  5. Insert an auxiliary variable into the problem TT if needed
  6. Go back to determine XX.
Detailed explanation

Practice Power Property of Logorithms

Test your knowledge with 6 quizzes

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

Examples with solutions for Power Property of Logorithms

Step-by-step solutions included
Exercise #1

2log38= 2\log_38=

Step-by-Step Solution

To solve this problem, let's simplify 2log382\log_3 8 using logarithm rules.

  • Step 1: Recognize the expression form
    The expression is of the form alogbca \cdot \log_b c, where a=2a = 2, b=3b = 3, and c=8c = 8.
  • Step 2: Apply the power property
    According to the power property of logarithms, 2log382 \cdot \log_3 8 can be simplified to log3(82)\log_3 (8^2).
  • Perform the calculation
    Calculate 828^2, which is 6464.
  • Step 3: Simplify further
    Therefore, we have log364\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 2log382\log_3 8 is log364\log_3 64.

Answer:

log364 \log_364

Video Solution
Exercise #2

3log76= 3\log_76=

Step-by-Step Solution

To simplify the expression 3log76 3\log_76 , we apply the power property of logarithms, which states:

alogbc=logb(ca) a\log_b c = \log_b(c^a)

Step 1: Identify the given expression: 3log76 3\log_76 .

Step 2: Apply the power property of logarithms:

3log76=log7(63) 3\log_76 = \log_7(6^3)

Step 3: Calculate 63 6^3 :

63=6×6×6=36×6=216 6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216

Step 4: Substitute back into the logarithmic expression:

log7(63)=log7216 \log_7(6^3) = \log_7216

Therefore, the simplified expression is log7216\log_7216.

Comparing with the answer choices, the correct choice is:

log7216 \log_7216

Answer:

log7216 \log_7216

Video Solution
Exercise #3

xln7= x\ln7=

Step-by-Step Solution

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

  • Step 1: Recognize that the expression xln7 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(ab)=blna\ln(a^b) = b \ln a. This tells us that if we have something of the form blna b \ln a , we can express it as ln(ab)\ln(a^b).
  • Step 3: Utilize the known expression and rule by substituting a=7 a = 7 and b=x b = x . Thus, xln7 x \ln 7 becomes ln(7x)\ln(7^x).

Therefore, the rewritten expression for xln7 x \ln 7 using logarithm rules is ln7x \ln 7^x .

This matches choice 4 from the provided options.

Answer:

ln7x \ln7^x

Video Solution
Exercise #4

log68= \log_68=

Step-by-Step Solution

To solve the problem log68 \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 23 2^3 , because 8 equals 2 multiplied by itself three times.

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

  • logb(an)=nlogba\log_b (a^n) = n \log_b a

Applying this property to log68\log_6 8, we have:

log68=log6(23)\log_6 8 = \log_6 (2^3)

Using the power property, this becomes:

log6(23)=3log62\log_6 (2^3) = 3 \log_6 2

Therefore, the expression for log68\log_6 8 in terms of log62\log_6 2 is:

3log623 \log_6 2.

Answer:

3log62 3\log_62

Video Solution
Exercise #5

nlogxa= n\log_xa=

Step-by-Step Solution

To solve this problem, we need to transform the expression nlogxa n\log_xa using the properties of logarithms.

  • Step 1: Identify the expression: We are given nlogxa n\log_xa , where logxa \log_xa is the logarithm of a a to the base x x , and n n is a coefficient.
  • Step 2: Use the power property of logarithms: The power property of logarithms states that if we have a logarithmic term multiplied by a coefficient n n , like nlogb(a) n\log_b(a) , it can be rewritten as logb(an) \log_b(a^n) .
  • Step 3: Apply the power property: By applying this property to nlogxa n\log_xa , we rewrite it as logx(an) \log_x(a^n) . This is because multiplying the logarithmic term by an external coefficient is equivalent to taking the argument a a to the power of that coefficient, n n .
  • Step 4: Conclusion about the transformation: This transformation demonstrates how the power property helps simplify expressions involving logarithms by turning multiplication into an exponentiation within the logarithm itself.

Therefore, the expression nlogxa n\log_xa can be transformed and expressed as logxan \log_xa^n by using the power property of logarithms.

Answer:

logxan \log_xa^n

Video Solution

Frequently Asked Questions

What is the power property of logarithms and when do I use it?

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The power property states that log_a(a^x) = x, which allows you to eliminate the logarithm and base when they match. Use this rule when solving equations where logarithms appear in exponents, as it simplifies complex expressions into manageable algebraic equations.

How do I solve equations with logarithms in the exponent?

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Follow these steps: 1) Take the logarithm with the same base on both sides, 2) Apply log_a(a^x) = x rule, 3) Create common bases using change of base formula, 4) Convert solvable logs to numbers, 5) Use auxiliary variables if needed, 6) Solve for the original variable.

Why do I need to use auxiliary variables when solving logarithmic equations?

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Auxiliary variables (like T = log_2 x) simplify complex logarithmic equations into familiar algebraic forms. This technique transforms equations with multiple logarithmic terms into quadratic or linear equations that are easier to solve and factor.

What logarithm rules do I need to know for power property problems?

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Essential rules include: • log_a(a^x) = x (power property) • log_a(xy) = log_a x + log_a y (product rule) • Change of base: log_a x = (log_b x)/(log_b a) • Basic conversions like log_2 4 = 2 and log_2 16 = 4

How do I check if my logarithm power property solution is correct?

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Substitute your answer back into the original equation and verify both sides are equal. Also check that your solution satisfies the domain restrictions (arguments of logarithms must be positive). If using auxiliary variables, ensure you've correctly converted back to the original variable.

What's the difference between solving log equations with variable bases vs numerical bases?

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Variable base equations (like x^(log expression) = number) require taking logs of both sides and using change of base formulas. Numerical base equations are simpler and often don't need auxiliary variables. The solution method is identical, but variable bases create more complex algebraic manipulations.

Common mistakes when applying power property of logarithms?

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Frequent errors include: 1) Forgetting to take logs of both sides initially, 2) Misapplying log_a(a^x) = x when bases don't match, 3) Calculation errors when converting logs to numbers, 4) Not checking domain restrictions, 5) Mixing up auxiliary variable substitutions when solving back for the original variable.

When should I use change of base formula in logarithm power problems?

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Use change of base when you have different logarithm bases in the same equation and need to create a common base for solving. Convert log_x expressions to a more convenient base (usually base 2 or 10) to simplify calculations and combine like terms effectively.

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