Intermediate Algebra/Logarithms

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Logarithms are used in several common numbering structures in the world. For instance, the pH scale is based on the base 10 logarithm. As a result, a pH of 5 is ten times more acidic than 6, and a pH of 4 is one hundred times more acidic than a pH of 6.

Logarithms

Syntax, or "what goes where", is one of the necessary evils that must be memorized in math. An exponential equation can be represented with letters like so: y=bx

But when put into logarithmic form, it looks like this:

logb(y)=x 

In the above example, b is the base, x is the exponent, and y is the product. Here's an example using constants:

log2(32)=5 

The above logarithmic equation can be written as an exponential equation like this:

25=32

Once, you remember that the base of the exponent is the number being raised to a power and that the base of the logarithm is the subscript after the log, the rest falls into place. I like to draw an arrow (either mentally or physically) from the base, to the exponent, to the product when changing from logarithmic form to exponential form. So visually or mentally I would go from 2 to 5 to 32 in the logarithmic example which (once I add the conventions) gives us: 25=32

So, when you are given a logarithm to solve, just remember how to convert it to an exponential equation. Here are some practice problems, the answers are at the bottom.

Properities of Logarithms

The following properities derive from the definition of logarithm.

If b>0 with b<>0, then for every real y,a,c it is:

1)logb(ya)=alogb(y) 

2)logb(ba)=a 

3)logb(ac)=logb(a)+logb(c) 

4)logb(a/c)=logb(a)logb(c) 


There is also the "change of base rule":

logb(a)=logd(a)logd(b) for any d<>0

Proof logb(a) is the same as bc=a.

Now, let us take the log to base d of both sides:

logd(bc)=logd(a).

Next, notice that the left side of this equation is the same as that in property number 1 above. Let us apply this property.

clogd(b)=logd(a)

Finally, solving this equation for c gives the desired result

c=logd(a)logd(b)


This rule allows you to evaluate logs to a base other than e or 10 on a calculator. For example, log3(12)=log10(12)log10(3)=2.262


  • Solve these logarithms
    • log3(81)=
    • log6(216)=
    • log4(64)=
  • Evaluate with a calculator
    • log4(6)=
  • Find the y value of these logarithms
    • log3(y)=3
    • log5(y)=4
    • log9(y)=4

Answers

  • Solve these logarithms
    • 4
    • 3
    • 3
  • Evaluate with a calculator
    • 1.2948
  • Find the y value of these logarithms
    • 27
    • 625
    • 6561

Logarithms are the reverse of exponential functions, just as division is the reverse of multiplication, for example:

5×6=30 and 30/6=5

73=343
log7343=3

Or, in a more general form, if ab=x, then logax=b. Also, if f(x)=ax, then f1(x)=logax, so if the two equations are graphed each is the reflection of the other over the line y=x. (in both equations, a is considered to be the base).

Because of this, alogab=b and logaab=b.

Common bases used are bases of 10 which is a called a common logarithm or e which is called a natural logarithm" (e~=2.71828182846).

Common logs are written either as log10x or simply as logx.

Natural logs are written either as logex or simply as lnx (the ln stands for natural logarithm).

Logarithms are commonly abbreviated as logs.

Properties of Logarithms

  1. logax+logay=logax*y
  2. logaxlogay=logaxy
  3. logaxb=b×logax

Proof:
logax+logay=logax*y

logax+logay

logax=b and logay=c

 ab=x and  ac=y

 xy=abac

 xy=a(b+c)

logaxy=b+c

and replace b and c (as above)

logaxy=logax+logay

Change of Base Formula

logyx=logaxlogay where a is any positive number. Generally, a is either 10 (for common logs) or e (for natural logs).

Proof:
logyx=a

 ya=x

Put both sides to loga

logaya=logax

 alogay=logax

 a=logaxlogay

Replace a from first line

logyx=logaxlogay

Swap of Base and Exponent Formula

alogbc=clogba where a or c must not be equal to 1.

Proof:

logab=1logba by the change of base formula above.

Note that a=clogca. Then

alogbc can be rewritten as

(clogca)logbc or by the exponential rule as

clogca*logbc

using the inverse rule noted above, this is equal to

clogca*1logcb

and by the change of base formula

clogba