Exponents and logarithms
We define what we mean by exponents:
b0 = 1
b1 = b
bn = b∙ b∙ b∙ b∙…∙ b
1
b-n = b∙ b∙ b∙ b∙…∙ b
𝑛
b1/n = √𝑏
𝑞
bp/q = √𝑏 𝑝
b multiplied by itself n times
1 divided by b n times
the n-th root of b
the q-th root of b to the p-th power
br = The limit of bp/q as p/q gets closer and closer to the real number r
And we note that we have some rules that go with exponents
1. bn∙bm = bn+m When multiplying the same base raised to powers, add the powers
2. (bn)m= bnm When raising a power to a power, multiply the powers.
Now let us define what we mean by logarithms:
Suppose that bu=v, that is, we get the result v when we raise b to the u power.
Then u is the exponent needed to raise base b to in order to obtain v.
This is an awkward statement so we make up a name for this.
u is the logarithm to the base b of y and we write this as u=Logby
So these two statements express the same relationship between the variables:
1. bu=v
2. u=Logby
You should feel free to move back and forth from exponential to logarithmic expression of the
relationship.
The rules for logarithms come from the rules for exponents:
Suppose that u=Logbx and that v=Logby
Then expressing these equations in exponential form gives us bu=x and bv=y.
Now consider that x∙y= bu∙ bv =bu+v so that Logb x∙y =u+v= Logbx+ Logby
Thus, from the exponent rule for adding exponents when we multiply bases raised to powers we get the
rule for logarithms: The logarithm of a product is the sum of the logarithms of the numbers being
multiplied. Logb x∙y = Logbx+ Logby
Similarly, raising a power to a power (bn)m= bnm gives us a rule for logarithms:
Let u=Logbx so that bu=x
xm =( bu)m=bum
Thus, Logbxm= Logb bum = um = mu = m Logbx.
We thus have the rule for taking logarithms that says the logarithm of x raised to a power n is n times
the logarithm of x
A summary of the rules:
1.
2.
3.
4.
5.
6.
bn∙bm = bn+m
(bn)m= bnm
Logb x∙y = Logbx+ Logby
Logbxm = m∙ Logbx
b0 = 1 gives us Logb1=0
b1 = b gives us Logbb=1
When multiplying the same base raised to powers, add the powers.
When raising a power to a power, multiply the powers.
The log of a product is the sum of the logs.
An exponent in a logarithm can move in front of the log as a multiplier.
The log of 1 is 0
The log of the base b is 1
Suppose that we know how to find logarithms to base 10 (LOG on the TI calculators) or logarithms to the
base e (LN on the TI calculator) but what we really need is log to the base 3 of some number. How
would we calculate this?
We want Log337 so let x= Log337 then in exponential form this is 3x=37
So let’s use our rules. Take the log of both sides to the base 10:
Log103x = Log1037
Using the logarithm rule for exponents gives us
Logb3x= x∙ Log103
And thus,
x∙ Log103 = Log1037
x=
Log10 37
Log10 3
but x was what we were trying to find x= Log337, so that
Log10 37
Log337=
Log10 3
This is an instance of the change of base formula for logarithms:
𝐋𝐨𝐠 𝐱
𝐋𝐨𝐠 𝒄 𝐱 = 𝐋𝐨𝐠𝒃 𝐜 We calculate the logarithm in base c, by knowing logarithms in base b.
𝒃
Why didn’t we have a change of base formula for exponentials? Well, you need logarithms for that.
Example: Express 34 as a power of 2.
Let x=34 and take the log to the base 2 of both sides:
Log2x = Log234= 4Log23
Now raise 2 to the powers Log2x and 4Log23
2Log2 𝑥 = 24Log2 3
But 2Log2 𝑥 = 𝑥 𝑎𝑛𝑑 𝑥 = 34
So 34 = 24Log2 3
The change of base formula for exponentials is:
𝒄𝒏 = 𝒃𝐧𝐋𝐨𝐠 𝒃𝒄 = 𝒃(𝐋𝐨𝐠 𝒃𝒄)𝒏
We could also arrive at this by noting that raising to a power and taking logarithms are inverse
operations:
𝑐 = 𝑏 Log𝑏 𝑐 and then raising both sides to the n-th power.
A summary of change of base formulas:
1.
𝐋𝐨𝐠 𝐱
𝐋𝐨𝐠 𝒄 𝐱 = 𝐋𝐨𝐠𝒃 𝐜
𝒃
𝒏
2. 𝒄 = 𝒃
𝐧𝐋𝐨𝐠 𝒃 𝒄
= 𝒃(𝐋𝐨𝐠 𝒃𝒄)𝒏