- XLog3base3 = Log3base7
Recall, Log3 base 3 = 1 (Log of the same base = 1)
X × 1 = Log3base7
X = Log3base7 ( Find Log 7 base 3 )
X = 1.78
(ii) 2^x + 2^x–¹ = 2
Solution
2^x + 1/2^x= 2 [ any number raise to power minus sign(-) is inverse of that number]
2^x + 1/2^X = 2 ( find l.c.m)
(4x² + 1) ÷2x =2 [ cross multiply ]
4² + 1 =2 × 2x
4x² + 1 = 4x
4x² – 4x + 1 =0
Using factorization method,
(4x² – 2x) –( 2x + 1) =0 [ hint: what’s common in the both side]
2x (2x – 1) -1(2x-1) = 0
(2x – 1)(2x-1) = 0
(2x – 1) appears twice
2x – 1 = 0
2x = 1 [ divide both side by 2]
X = ½(twice)
To prove, put X as ½ into the real equation
2^X + 1/2^x = 2
2^½+ 1/2^½= 2
1.414 + 0.707 = 2 ( approximately)
b. simplify Log1/5125 (ii) Prove that Logba × Logcb × Logac = 1
solution
C. If Log2 base5= 0.431 and Log3base 5 =0.682, evaluate Log54/9
Solution
evaluate Log54/9 = Log54 ÷ Log59
Log522 ÷ Log532 = 2Log52 ÷ 2Log53
(2 × 0.431) ÷ ( 2 × 0.682)
0.862/1.364 = 0.632( approximately)
Log1/5125 = Log5-1 125
Let the both be in base 5
Log5125 ÷ Log55-1
Log553 ÷ Log55-1
3Log55 ÷ -1Log55
3÷-1 = -3
(ii) Prove that Logba × Logcb × Logac = 1
We’ll also apply the same thing to this, only that their bases will be different
Logba × Logcb × Logac
(Logaa ÷ Logbb) × (Logbb ÷ Logcc) ×( Logcc ÷ Logaa)
As we know Log of the same base = 1
1/1 × 1/1 × 1/1 = 1 × 1 × 1
=1
Question on surd
2-1/(√2-1)
Using rationalization method,
2-1/(√2-1) × (√2+1)/(√2+1)
2- (√2+1)/ (2+√2-√2-1)
2-(√2+1)/1
2-√2-1
1-√2
