Solution To Exercises in Logarithm
The following exercises have been extracted from the New General Mathematics for Senior Secondary Schools Book 3.
Solve the Following equations for x.
(a) log10x=3
We simply use the basic rule of logarithm to solve the question which is P=logbN⇒N=bP
=103=x
⇒x=1000
(b) logx27=3
⇒x3=27
⇒x3=33
x=3
(c) 3x=1
Remember from the laws of indices a0=1
3x=30
x=0
(d) 4x+1=8x
we simply equate the bases, so that we can cancel out the bases
⇒22(x+1)=23x
cancel out the bases
2x+2=3x
collect like terms and make x subject of formula
x=2
(e) 2x×23x−1=1
by the law addition law of indices, this will become
2x+(3x−1)=1
remeber a0=1
24x−1=20
cancel out the bases
4x−1=0
4x=1
x=14
(f) 22x−3×2x+2=0
We collect like terms and use law of indices to simplify bases
22x+x−3+2=0
23x−1=0
23x=1
23x=20
3x=0
x=0
(g) 6log(x+4)=log64
Using the power law of Logarithm
log(x+4)6=log26
(x+4)6=26
x+4=2
x=−2
(h) 92x+1=81x−23x
We simply equate the bases of the L.H.S and R.H.S
32(2x+1)=34(x−2)3x
34x+2=33x−8
4x+2=3x−8
x=−10
(i) log10(2x+1)−log10(3x−2)=1
Remember log101=1
log10(2x+1)−log10(3x−2)=log10
log10(2x+13x−2)=log10
2x+13x−2=10
Cross multiply
2x+1=10(3x−2)
2x+1=30x−20
2x−30x+20+1=0
−28x+21=0
−28x=−21
x=2128=34
(j) 3×91+x=27−x
Simplify the bases in such a way that they are all the same
3×32(1+x)=3−3x
31+2+2x=3−3x
Cancel out the bases
1+2+2x=−3x
Collect like terms
3=−3x−2x
3=−5x
x=−35
(k) log10(3x−1)−log102=3
Remember log103=log1000=3
log10(3x−12)=log103
3x−12=1000
Cross Multiply
3x−1=2000
Collect like terms
3x=1999
x=19993
x=666.3
The following exercises have been extracted from the New General Mathematics for Senior Secondary Schools Book 3.
Solve the Following equations for x.
(a) log10x=3
We simply use the basic rule of logarithm to solve the question which is P=logbN⇒N=bP
=103=x
⇒x=1000
(b) logx27=3
⇒x3=27
⇒x3=33
x=3
(c) 3x=1
Remember from the laws of indices a0=1
3x=30
x=0
(d) 4x+1=8x
we simply equate the bases, so that we can cancel out the bases
⇒22(x+1)=23x
cancel out the bases
2x+2=3x
collect like terms and make x subject of formula
x=2
(e) 2x×23x−1=1
by the law addition law of indices, this will become
2x+(3x−1)=1
remeber a0=1
24x−1=20
cancel out the bases
4x−1=0
4x=1
x=14
(f) 22x−3×2x+2=0
We collect like terms and use law of indices to simplify bases
22x+x−3+2=0
23x−1=0
23x=1
23x=20
3x=0
x=0
(g) 6log(x+4)=log64
Using the power law of Logarithm
log(x+4)6=log26
(x+4)6=26
x+4=2
x=−2
(h) 92x+1=81x−23x
We simply equate the bases of the L.H.S and R.H.S
32(2x+1)=34(x−2)3x
34x+2=33x−8
4x+2=3x−8
x=−10
(i) log10(2x+1)−log10(3x−2)=1
Remember log101=1
log10(2x+1)−log10(3x−2)=log10
log10(2x+13x−2)=log10
2x+13x−2=10
Cross multiply
2x+1=10(3x−2)
2x+1=30x−20
2x−30x+20+1=0
−28x+21=0
−28x=−21
x=2128=34
(j) 3×91+x=27−x
Simplify the bases in such a way that they are all the same
3×32(1+x)=3−3x
31+2+2x=3−3x
Cancel out the bases
1+2+2x=−3x
Collect like terms
3=−3x−2x
3=−5x
x=−35
(k) log10(3x−1)−log102=3
Remember log103=log1000=3
log10(3x−12)=log103
3x−12=1000
Cross Multiply
3x−1=2000
Collect like terms
3x=1999
x=19993
x=666.3