- Properties of Logarithms,
Exponential Logarithmic Forms
Property.
For example, log base 2[8] = 3,
since 2 to the 3rd power = 8.
The equation log base 2[8] = 3
is in logarithmic form.
The equation 2 to the 3rd power
= 8 is in exponential form.
The forms log base 2[8] = 3
and 2 to the 3rd power = 8
are equivalent.
Here are more examples
of equations
in equivalent logarithmic
and exponential forms.
The logarithmic form
log base 2 [16] = 4
is the same as
2 to the 4th power = 16.
The logarithmic form
of log base 3[9] = 2
is the same as 3 squared = 9.
The logarithmic form
of log base 5[125] = 3
is the same as the exponential
form 5 to the 3rd = 125.
The logarithmic form
of log of 100,000 = 5
is the same as the exponential
form,
10 to the 5th power = 100,000.
Properties of Logarithms,
the exponential logarithmic
forms property.
For A greater than 0, B greater
than 0, and B not equal to 1,
the equations log base
B[A] = C and B to the C = A
are equivalent.
The equation log base B[A] = C
is in logarithmic form
and the equation
B to the C power = A
is in exponential form.
Either form can replace the
other when you solve a problem.
In this example,
we want to write the equation
in exponential form.
Assume all constants are
positive and not equal to 1.
Part A, I have
log base 5[125] = 3.
To rewrite this, first,
we find our base which is 5,
the exponent
is what it's equal to,
so I have 5 to the 3rd equals,
and it is equal to what's in
the parentheses which is 125.
So 5 to the 3rd power = 125.
For part B, I have log base
K [T] = W.
So, I start with my base which
is K, its equal to the exponent,
so that's K to the W power,
and it's equal to what's
in the parentheses which is T.
So, log base K[T] = W is the
same as K to the W power = T.
In this example, write the
equation in logarithmic form.
Assume that all constants are
positive and not equal to 1.
Part A, I have
3 to the 4th power = 81.
So I start by writing log,
and then my small number
is always the base
which in this case is 3,
I put in parentheses what it
should equal which is 81,
and then that equals my exponent
which is 4.
So 3 to the 4th power = 81 is
the same as log base 3[81] = 4.
For part B,
I have P to the R power = Q.
So, I start with my log,
my base is P, my parentheses,
I put what it equals,
which is Q,
and it equals the exponent
which is R.
So P to the R power = Q
is the same as log base P[Q]
is the same as log base
P[Q] = R.
Logarithmic Equations,
a logarithmic equation
in one variable is an equation
in one variable that contains
one or more logarithms.
In summary, for an equation
of the form log base B[X] = K,
we can solve for B or X
by writing the equation
in exponential form.
In this example, we want to
solve log base 4[X] = 2.
My first step is to rewrite this
in exponential form.
So I start with my base
which is 4,
it's equal to the exponent 2
which is equal to what's
in parentheses X.
4 squared is 16, so X = 16.
In this example, solve
log[X] = -2,
since the base isn't written,
we know that the base is 10
is equal to my exponent -2,
10 to the -2 power = X, since X
is in the parentheses.
10 to the -2 power is 1/100,
so X = 1/100.
In this example, solve 4
x log base 81[X] - 3 = -2.
Our first step is to isolate
log base 81[X],
that way we can transform it.
So, I'm going to add 3
to both sides,
4 x log base 81[X] = 1
divide both sides by 4,
log base 81[X] = 1/4.
Now we're ready to rewrite it
in exponential form.
My base is 81, my power is 1/4,
and that's going to equal x.
Calculating, I have 81
to parenthesis 1 divided by 4,
power, enter, and I get 3,
so X is 3.
Power Property for Logarithms,
an exponential equation
in one variable
is an equation in one variable
in which an exponent contains
a variable.
Power Property for Logarithms.
If x is greater than 0,
b is greater than 0,
and b does not equal 1.
Log base B[X] to the P power
= P x log base B[X].
In words,
a logarithm of a power of X
is the exponent
x the logarithm of X.
In this example, solve for B.
Round any approximate solutions
to the fourth decimal place.
Log base B[8] = 3.
One method we have to solve
this is by remembering that 8
is a perfect cube.
So I have log base B
of 2 to the 3rd power = 3.
I can use my power property
to rewrite this as
3 x log base B[2] = 3
divide both sides by 3,
I have log base B[2] = 1.
I can rewrite this
in exponential form,
my base to the 1st power
= 2 which means B = 2.
In this example, solve for B.
Round any approximate solutions
to the fourth decimal place.
Log base B[16] = 5.
16 can be written as
2 to the 4th power
which gives us log base
B[2] to the 4th power = 5,
so I can rewrite my log using
the power property,
4 x log base B[2] = 5
dividing both sides by 4.
Log base B[2] = 5/4.
Now, I can rewrite that
in exponential form,
B to the 5/4 power = 2.
In order to solve this, we have
to get rid of our power.
I can get rid of any power
by taking it to its reciprocal
power.
So B shall be to the 5/4th power
to the 4/5th power
will give me B to the 20/20
which is B to the 1st
which is just B.
What I do to one side,
I have to do to the other,
so I have to find 2 to the 4/5th
power.
Calculating 2, exponent,
[4 divided by 5]
gives us 1.741101127,
we want to round this to four
decimal places.
So B = 1.7411.
Logarithmic Property
of Equality,
for positive real numbers,
A, B, and C,
where B does not equal 1,
the equations A = C,
and log base B[A]
= log base B[C] are equivalent.
In this example,
we want to solve,
round the solution to the fourth
decimal place.
5 to the X power = 97.
Based on the logarithmic
property of equality,
I can take the log of both sides
and it still remain equivalent,
because I want to be able
to calculate this,
I'm going to use log base 10.
So, log of 5 to the X = log[97].
I can write my exponent
in front, X x log[5] = log[97].
Now, I can solve for X by
dividing both sides by log[5],
X = log[97]/log[5],
calculating log[97],
make sure you close your
parenthesis, divided by log[5],
close your parenthesis, enter,
2.842427746,
rounding to the fourth decimal
place, 2.8424.
So X = 2.8424.
In this example, we want to
solve and round the solution
to the fourth decimal place.
3 to the 4X + 2 power = 341.
First step,
take log of both sides.
Log of 3 to the 4X + 2 power
= log[341].
Now, we're ready to bring
our exponent in front
using our power property
which gives us
[4X + 2] x log[3] = log[341].
Now we can divide both sides
by log[3],
so we have 4X + 2 = log[341]
divided by log[3].
Calculating log[341],
close parenthesis,
divided by log[3] which gives us
5.3084 or X + 2 = 5.3084.
Subtract both sides by 2
or X = 3.3084,
dividing by 4, X = 0.8271.
So for the equation, 3 to the 4X
+ 2 power = 341, X = 0.8271.
In this example, use intersect
on a graphing calculator
to solve the equation.
Round any solutions
to the fourth decimal place.
2 to the X = 5 - 2X.
What we want to do is look
at this as Y = 2 to the X power
and Y = 5 - 2X and type those
two equations in our calculator.
Y equals,
we want to type in 2 exponent X,
arrow down,
and then type 5 - 2X.
Now, we want to go to zoom,
and let's choose six standard.
Our blue line is 2 to the X
power, our green line is 5 - 2X.
As long as we can see
where they intersect,
we'll be able to use
our intersect function
on the calculator.
Second trace gets me
to my calculate menu,
I want number five and an arrow
over to where they cross, enter,
enter again, enter, and my
intersection X = 1.2831506,
Y = 2.4336988.
We're interested in one axis,
using our fourth decimal place,
X = 1.2832.
Properties of Logarithms,
I can write log base B[A] = C,
as B to the C power equals A,
that's our Exponential
Logarithmic Forms Property.
I can use this property to solve
my logarithmic equations.
If my X is in the position of A,
all I have to do is rewrite
the logarithmic form
and exponential form
and then I can solve for X.
If my X is in the position
of my base,
I can also rewrite it as
exponential form and solve.
The power property for Logarithm
says that
if I have log base B[X]
to the P power,
I can rewrite that as
P = log base B[X].
The Logarithm Property
of Equality says that
I can take the logarithm
of any base
to each side of the equation
to solve
using log of base 10
allows us to use our calculator
to assist us in solving.
