Change of Logarithm Base

🏆Practice change-of-base formula for logarithms

Change of Logarithm Base

Reminder - Logarithms

The definition of the logarithm:
logax=blog_a⁡x=b
X=abX=a^b

Where:
aa is the base of the log
XX is what appears inside the log - can also appear inside of parentheses
bb is the exponent to which we raise the base of the log in order to obtain the number that appears inside of the log.

How to change the base of a logarithm?

According to the following rule:
logaX=logthe base we want to change toXlogthe base we want to change toalog_aX=\frac{log_{the~base~we~want~to~change~to}X}{log_{the~base~we~want~to~change~to}a}

Logarithmic change of base formula illustrated: log base b of a equals log base x of a divided by log base x of b, with arrows showing transformation from original form.

In the numerator there will be a log with the base we want to change to, as well as what appears inside of the original log.
In the denominator there will be a log with the base we want to change to, and the content will be the base of the original log.

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Test yourself on change-of-base formula for logarithms!

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\( \frac{\log_85}{\log_89}= \)

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Change of Logarithm Base

Logarithm - Reminder

First, let's recall the definition of the log and understand what is the base of a logarithm:

The definition of the logarithm is:
logax=blog_a⁡x=b
X=abX=a^b

Where:
aa is the base of the log
XX is what appears inside the log - which can also appear inside of parentheses
bb is the exponent to which we raise the base of the log to in order to obtain the number that appears inside of the log.

The rule states that if we raise base aa to the power of bb we obtain XX.
In order to solve a log, we ask ourselves - to what power should we raise aa to obtain XX.
The power bb that we found is the solution.

For example:
When we have the expression
log525=log_5⁡25=

Let's ask ourselves –
To what power do we need to raise 55 to in order to obtain 2525? The answer is 22. Therefore:
log525=2log_5⁡25=2


Important to know - the log in the example is read as follows:
log base 55 of 2525

Change of logarithm base

You're probably asking yourself, why do we need to change the base of the logarithm?
Excellent question!
Sometimes in subtraction, addition, multiplication, or division exercises with different bases, it is easier for us to change the base of one of the logarithms so we can use the addition and subtraction formulas, and the same goes for multiplication and division.
So how do we do it? According to the following rule:

The Change of Base Rule for Logarithms:

logaX=logbase we want to change toXlogbase we want to change toalog_aX=\frac{log_{base~we~want~to~change~to}X}{log_{base~we~want~to~change~to}a}

Let's look at an example:
Here is the exercise:
log816=log_8⁡16=
Convert it to a log with base 22.

Solution:
Upon observation this is an illogical solution... To what power should we raise 88 to in order to obtain 1616?
Therefore, we can change the base of the log and solve the problem far more easily! Note that the instruction is to change the log to base 22.
To change the base of the log, we'll use the formula:
In the numerator, we'll have log base 22 (the base we want to change to) with the original log content 1616
And in the denominator, we'll have log base 22 (the base we want to change to) with 88 the original base.
We obtain the following:
log816=log216log28log_8⁡16=\frac{log_2⁡16}{log_2⁡8 }
Are we done? Not yet. Now we must proceed to solve the logs until we obtain a number.
log216=4log_2⁡16=4
log28=3log_28=3
We insert the data into the exercise as seen below to obtain our solution:
=43=\frac{4}{3}
Amazing!

Now we'll increase the difficulty level and move on to another exercise with a variable where we'll also use the change of logarithm base rule:
log6x+log36x=3log_6⁡x+log_{36}⁡x=3

Solution
First of all, don't panic. With the rule you just learned, you can solve this exercise very simply. Let's start step by step.

When encountering two logarithms with different bases, the first step is to convert the larger log to the smaller one because it will be simpler to solve.
Let's do it - we'll convert log36xlog_{36}⁡x to base 66 and get:
log36x=log6xlog636log_{36}⁡x=\frac{log_6x}{log_6⁡36 }

Insert the data into the exercise as follows:

log6x+log6xlog636=3log_6⁡x+\frac{log_6⁡x}{log_6⁡36} =3

Let's continue solving the exercise.
We can solve the denominator -
log636=3log_6⁡36=3
Insert into the exercise
log6x+log6x2=3log_6⁡x+\frac{log_6⁡x}{2}=3〗
Let's continue solving the problem and write the exercise in a simpler form:

log6x+0.5log6x=3log_6⁡x+0.5 log_6⁡x=3
Let's combine like terms:
1.5log6x=31.5 log_6x=3

log6x=2log_6 x=2
Solve the following –
x=62x=6^2
x=36x=36

Tip - Sometimes it's useful to substitute an auxiliary variable tt instead of the entire log.
When reached this stage in the exercise -
log6x+log6x2=3log_6⁡x+\frac{log_6⁡x}{2}=3
We can substitute
log6x=tlog_6 x=t
as follows:
t+t2=3t+\frac{t}{2}=3
1.5t=31.5t=3
t=2t=2

But notice, this is not the result. Now we need to substitute 22 instead of tt
and determine xx
we obtain the following solution :
log6x=2log_6 x=2
62=x6^2=x
x=36x=36

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Examples with solutions for Change-of-Base Formula for Logarithms

Exercise #1

log85log89= \frac{\log_85}{\log_89}=

Video Solution

Step-by-Step Solution

To solve this problem, let's simplify the given expression log85log89\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 logbMlogbN=logNM\frac{\log_b M}{\log_b N} = \log_N M.
  • Step 3: Apply the division rule to the given expression: log85log89=log95\frac{\log_8 5}{\log_8 9} = \log_9 5.

Thus, after simplifying, we see that log85log89=log95\frac{\log_85}{\log_89} = \log_9 5.

Hence, the correct answer is log95\log_9 5, which corresponds to the choice 1.

Answer

log95 \log_95

Exercise #2

log74= \log_74=

Video Solution

Step-by-Step Solution

To solve the problem of evaluating log74\log_7 4, we will use the change-of-base formula for logarithms.

The change-of-base formula is:

  • logba=logkalogkb\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:

log74=ln4ln7\log_7 4 = \frac{\ln 4}{\ln 7}

By applying the change-of-base formula, we find that the logarithm log74\log_7 4 can be expressed as ln4ln7\frac{\ln 4}{\ln 7}.

Upon examining the provided choices, we identify that choice 2: ln4ln7\frac{\ln 4}{\ln 7} matches our result.

Therefore, the solution to the problem is ln4ln7\frac{\ln 4}{\ln 7}.

Answer

ln4ln7 \frac{\ln4}{\ln7}

Exercise #3

log4x9log4xa= \frac{\log_{4x}9}{\log_{4x}a}=

Video Solution

Step-by-Step Solution

To solve the given expression log4x9log4xa\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: log4x9=loga9loga(4x)\log_{4x}9 = \frac{\log_a 9}{\log_a (4x)} and log4xa=logaaloga(4x)\log_{4x}a = \frac{\log_a a}{\log_a (4x)}.
  • Step 2: Substitute these into our original expression:
    log4x9log4xa=loga9loga(4x)logaaloga(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 loga(4x)\log_a (4x) cancels out from the numerator and the denominator, leaving us with loga9logaa\frac{\log_a 9}{\log_a a}.
  • Step 4: Further simplify using the fact that logaa=1\log_a a = 1 because any number aa to the power of 1 is aa.
    This results in loga91=loga9\frac{\log_a 9}{1} = \log_a 9.

Therefore, the expression simplifies to loga9\log_a 9.

The correct answer is loga9\log_a 9, which matches choice 1.

Answer

loga9 \log_a9

Exercise #4

log9e2log9e= \frac{\log_9e^2}{\log_9e}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll simplify the given expression log9e2log9e\frac{\log_9e^2}{\log_9e} using logarithmic rules:

Step 1: Apply the power rule of logarithms:
The numerator log9e2\log_9e^2 can be rewritten using the power rule: log9e2=2log9e\log_9e^2 = 2 \cdot \log_9e.

Step 2: Substitute and simplify the fraction:
Substitute the expression from Step 1 into the original problem:
log9e2log9e=2log9elog9e\frac{\log_9e^2}{\log_9e} = \frac{2 \cdot \log_9e}{\log_9e}.

Step 3: Cancel common terms:
Since log9e\log_9e appears in both the numerator and the denominator, it cancels out, leaving:
2log9elog9e=2 \frac{2 \cdot \log_9e}{\log_9e} = 2 .

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

Answer

2 2

Exercise #5

log89alog83a= \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 log89alog83a\frac{\log_8 9a}{\log_8 3a}, we can directly apply the quotient rule for logarithms, which tells us that logbMlogbN=logNM\frac{\log_b M}{\log_b N} = \log_N M.
Step 2: Applying this formula, we find that log89alog83a=log3a9a\frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a.

Therefore, the solution to the problem is log3a9a \log_{3a} 9a .

Answer

log3a9a \log_{3a}9a

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