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}

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.

Suggested Topics to Practice in Advance

  1. Addition of Logarithms
  2. Power in logarithm
  3. Changing the Base of a Logarithm

Practice Change-of-Base Formula for Logarithms

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

Exercise #1

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

Video Solution

Answer

log95 \log_95

Exercise #2

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

Video Solution

Answer

loga9 \log_a9

Exercise #3

log89alog83a= \frac{\log_89a}{\log_83a}=

Video Solution

Answer

log3a9a \log_{3a}9a

Exercise #4

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

Video Solution

Answer

2 2

Exercise #5

log74= \log_74=

Video Solution

Answer

ln4ln7 \frac{\ln4}{\ln7}

Exercise #6

ln4x= \ln4x=

Video Solution

Answer

log74xlog7e \frac{\log_74x}{\log_7e}

Exercise #7

Is inequality true?

\log_{\frac{1}{4}}9<\frac{\log_57}{\log_5\frac{1}{4}}

Video Solution

Answer

Yes, since:

\log_{\frac{1}{4}}9<\log_{\frac{1}{4}}7

Exercise #8

2log7(x+1)log7e=ln(3x2+1) \frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1)

x=? x=\text{?}

Video Solution

Answer

1,0 1,0

Exercise #9

log4(x2+8x+1)log48=2 \frac{\log_4(x^2+8x+1)}{\log_48}=2

x=? x=\text{?}

Video Solution

Answer

4±79 -4\pm\sqrt{79}

Exercise #10

Find X

log84x+log8(x+2)log83=3 \frac{\log_84x+\log_8(x+2)}{\log_83}=3

Video Solution

Answer

1+312 -1+\frac{\sqrt{31}}{2}

Exercise #11

log45+log423log42= \frac{\log_45+\log_42}{3\log_42}=

Video Solution

Answer

log810 \log_810

Exercise #12

What is the domain of X so that the following is satisfied:

\frac{\log_{\frac{1}{8}}2x}{\log_{\frac{1}{8}}4}<\log_4(5x-2)

Video Solution

Answer

\frac{2}{3} < x

Exercise #13

log311log34+1ln32log3= \frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3=

Video Solution

Answer

log411+loge2 \log_411+\log e^2

Exercise #14

2log78log74+1log43×log29= \frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=

Video Solution

Answer

7 7

Exercise #15

3(ln4ln5log57+1log65)= -3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})=

Video Solution

Answer

3log5724 3\log_5\frac{7}{24}

Topics learned in later sections

  1. Logarithms