Change-of-Base Formula for Logarithms: Applying the formula

Examples with solutions for Change-of-Base Formula for Logarithms: Applying the formula

Exercise #1

log⁡85log⁡89= \frac{\log_85}{\log_89}=

Video Solution

Step-by-Step Solution

To solve this problem, let's simplify the given expression log⁡85log⁡89\frac{\log_85}{\log_89}.

  • Step 1: Recognize that both the numerator and denominator have the same base, 8.
  • Step 2: The division property of logarithms states that log⁡bMlog⁡bN=log⁡NM\frac{\log_b M}{\log_b N} = \log_N M.
  • Step 3: Apply the division rule to the given expression: log⁡85log⁡89=log⁡95\frac{\log_8 5}{\log_8 9} = \log_9 5.

Thus, after simplifying, we see that log⁡85log⁡89=log⁡95\frac{\log_85}{\log_89} = \log_9 5.

Hence, the correct answer is log⁡95\log_9 5, which corresponds to the choice 1.

Answer

log⁡95 \log_95

Exercise #2

log⁡74= \log_74=

Video Solution

Step-by-Step Solution

To solve the problem of evaluating log⁡74\log_7 4, we will use the change-of-base formula for logarithms.

The change-of-base formula is:

  • log⁡ba=log⁡kalog⁡kb\log_b a = \frac{\log_k a}{\log_k b}, where kk can be any base, commonly chosen as 10 (common logs) or ee (natural logs).

We will choose natural logarithms (ln⁡\ln) for simplicity, therefore:

log⁡74=ln⁡4ln⁡7\log_7 4 = \frac{\ln 4}{\ln 7}

By applying the change-of-base formula, we find that the logarithm log⁡74\log_7 4 can be expressed as ln⁡4ln⁡7\frac{\ln 4}{\ln 7}.

Upon examining the provided choices, we identify that choice 2: ln⁡4ln⁡7\frac{\ln 4}{\ln 7} matches our result.

Therefore, the solution to the problem is ln⁡4ln⁡7\frac{\ln 4}{\ln 7}.

Answer

ln⁡4ln⁡7 \frac{\ln4}{\ln7}

Exercise #3

log⁡4x9log⁡4xa= \frac{\log_{4x}9}{\log_{4x}a}=

Video Solution

Step-by-Step Solution

To solve the given expression log⁡4x9log⁡4xa\frac{\log_{4x}9}{\log_{4x}a} using the change-of-base formula, follow these steps:

  • Step 1: Apply the change-of-base formula to both the numerator and the denominator expressions.
    This gives us: log⁡4x9=log⁡a9log⁡a(4x)\log_{4x}9 = \frac{\log_a 9}{\log_a (4x)} and log⁡4xa=log⁡aalog⁡a(4x)\log_{4x}a = \frac{\log_a a}{\log_a (4x)}.
  • Step 2: Substitute these into our original expression:
    log⁡4x9log⁡4xa=log⁡a9log⁡a(4x)log⁡aalog⁡a(4x)\frac{\log_{4x}9}{\log_{4x}a} = \frac{\frac{\log_a 9}{\log_a (4x)}}{\frac{\log_a a}{\log_a (4x)}}.
  • Step 3: Simplify the fraction:
    The log⁡a(4x)\log_a (4x) cancels out from the numerator and the denominator, leaving us with log⁡a9log⁡aa\frac{\log_a 9}{\log_a a}.
  • Step 4: Further simplify using the fact that log⁡aa=1\log_a a = 1 because any number aa to the power of 1 is aa.
    This results in log⁡a91=log⁡a9\frac{\log_a 9}{1} = \log_a 9.

Therefore, the expression simplifies to log⁡a9\log_a 9.

The correct answer is log⁡a9\log_a 9, which matches choice 1.

Answer

log⁡a9 \log_a9

Exercise #4

log⁡9e2log⁡9e= \frac{\log_9e^2}{\log_9e}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll simplify the given expression log⁡9e2log⁡9e\frac{\log_9e^2}{\log_9e} using logarithmic rules:

Step 1: Apply the power rule of logarithms:
The numerator log⁡9e2\log_9e^2 can be rewritten using the power rule: log⁡9e2=2⋅log⁡9e\log_9e^2 = 2 \cdot \log_9e.

Step 2: Substitute and simplify the fraction:
Substitute the expression from Step 1 into the original problem:
log⁡9e2log⁡9e=2⋅log⁡9elog⁡9e\frac{\log_9e^2}{\log_9e} = \frac{2 \cdot \log_9e}{\log_9e}.

Step 3: Cancel common terms:
Since log⁡9e\log_9e appears in both the numerator and the denominator, it cancels out, leaving:
2⋅log⁡9elog⁡9e=2 \frac{2 \cdot \log_9e}{\log_9e} = 2 .

Therefore, the solution to the problem is 2\boxed{2}.

Answer

2 2

Exercise #5

log⁡89alog⁡83a= \frac{\log_89a}{\log_83a}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Apply the mathematical property using the quotient rule for logarithms.
  • Step 2: Simplify the expression into the desired format.

Now, let's work through each step:
Step 1: Given the expression log⁡89alog⁡83a\frac{\log_8 9a}{\log_8 3a}, we can directly apply the quotient rule for logarithms, which tells us that log⁡bMlog⁡bN=log⁡NM\frac{\log_b M}{\log_b N} = \log_N M.
Step 2: Applying this formula, we find that log⁡89alog⁡83a=log⁡3a9a\frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a.

Therefore, the solution to the problem is log⁡3a9a \log_{3a} 9a .

Answer

log⁡3a9a \log_{3a}9a

Exercise #6

ln⁡4x= \ln4x=

Video Solution

Step-by-Step Solution

To solve this problem, we’ll follow these steps:

  • Step 1: Identify the expression ln⁡4x\ln 4x.
  • Step 2: Apply the change-of-base formula to transform the natural logarithm to one using base 7.
  • Step 3: Use the formula ln⁡a=log⁡balog⁡be\ln a = \frac{\log_b a}{\log_b e} to substitute the values.

Now, let's work through each step:
Step 1: The expression given is ln⁡4x\ln 4x.
Step 2: We want a base 7 logarithm, so we apply the change-of-base formula:
Step 3: We have:

ln⁡4x=log⁡74xlog⁡7e\ln 4x = \frac{\log_7 4x}{\log_7 e}

Therefore, the logarithmic expression ln⁡4x\ln 4x in base 7 is equivalent to log⁡74xlog⁡7e\frac{\log_7 4x}{\log_7 e}.

This matches the correct answer choice, which is choice 4.

Answer

log⁡74xlog⁡7e \frac{\log_74x}{\log_7e}