Required Prior Knowledge
Questions
Calculate
a) 6a3b2×2a8b3
b) 10x9y÷5x3y7
c) 5x12(x−12−2x32)
d) √916x2
e) 16x+1+20×42x2x−3×8x+2
State the Laws of Indices.
Solutions
Get Ready
Question
Read through this proof. At each stage think carefully about what is being said and what is shows.
Let x=logbm
This implies that bx=m
Similarly y=logbn⇒by=n
Hence logbmn=logbbxby=logbbx+y=x+y=logbm+logbn
Thus logbm+logbn=logbmn
Explanation
Notes
There are four Laws of Logarithms which derive from the Laws of Indices.
The first of these is proven in the Get Ready.
Think about what the result for the other three might be and then check.
a) logbm+logbn=logbmn
b) logbm−logbn=
c) klogbm=
d) −logbm=
-
a) logbm+logbn=logbmn
b) logbm−logbn=logbmn
c) klogbm=logbmk
d) −logbm=logb1m
Use the proof in the Get Ready to construct similar proofs for the other three Laws of Logarithms.
Examples and Your Turns
Example
Express loga5+2loga7−loga35 as a single logarithm.
Your Turn
Express logap+2logaq−3logar as a single logarithm.
Your Turn
Express 1−logaab as a single logarithm.
Your Turn
Evaluate 2(log5+log2)−1
Your Turn
Simplify 2log54+3
Your Turn
Simplify log8log4
Your Turn
In each of these groups of three expressions, two are equal. Determine the odd one out, and write a fourth expression to match the value of the odd one out.
a) log327,log416,log5125
b) log40.25,log50.2,log100.01
c) log2x4+log2x3,log2x7,log2x2+log2x6
d) −2log5x,log5x3−log5x5,log5√x
-
log327=3log416=2log5125=3
Hence the odd one out is log416.
Possible matching values could be log39,log24,logxx2,...
-
log40.25=−1log50.2=−1log100.01=−2
Hence the odd one out is log100.01.
Possible matching values are log20.25,log0.1100,logx1x2,...
-
log2x4+log2x3=log2x12log2x7log2x2+log2x6=log2x12
Hence the odd one out is log2x7.
Possible matching values could be log2x+log2x6,log2x12−log2x5,...
-
−2log5x=log5x−2log5x3−log5x5=log5x−2log5√x=log5x12
Hence the odd one out is log5√x.
A possible matching value is 12log5x
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.