**XLog3base3 = Log3base7**

Recall, Log3 base 3 = 1 (Log of the same base = 1)

X ร 1 = Log3base7

X = Log3base7 ( Find Log 7 base 3 )

X = 1.78**(ii) 2^x + 2^xโยน = 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)

(4xยฒ + 1) รท2x =2 [ cross multiply ]4ยฒ + 1 =2 ร 2x

4xยฒ + 1 = 4x

4xยฒ โ 4x + 1 =0**Using factorization method,**

(4xยฒ โ 2x) โ( 2x + 1) =0 [ hint: whatโs common in the both side]2x (2x โ 1) -1(2x-1) = 0

(2x โ 1)(2x-1) = 0

(2x โ 1) appears twice

2x โ 1 = 0

2x = 1 [ divide both side by 2]X = ยฝ(twice)

**To prove, put X as ยฝ into the real equation**

2^X + 1/2^x = 2

2^ยฝ+ 1/2^ยฝ= 2

1.414 + 0.707 = 2 ( approximately)

## b. simplify Log1/5125 (ii) Prove that Logba ร Logcb ร Logac = 1

solution

C. If Log2 base5= 0.431 and Log3base 5 =0.682, evaluate Log54/9

Solution

evaluate Log54/9 = Log54 รท Log59

Log522 รท Log532 = 2Log52 รท 2Log53

(2 ร 0.431) รท ( 2 ร 0.682)

0.862/1.364 = 0.632( approximately)

Log1/5125 = Log5-1 125

Let the both be in base 5

Log5125 รท Log55-1

Log553 รท Log55-1

3Log55 รท -1Log55

3รท-1 = -3

(ii) Prove that Logba ร Logcb ร Logac = 1

Weโll also apply the same thing to this, only that their bases will be different

Logba ร Logcb ร Logac

(Logaa รท Logbb) ร (Logbb รท Logcc) ร( Logcc รท Logaa)

As we know Log of the same base = 1

1/1 ร 1/1 ร 1/1 = 1 ร 1 ร 1

=1

**Question on surd**

2-**1/(โ2-1)Using rationalization method,**

2-1/(โ2-1) ร (โ2+1)/(โ2+1)

2- (โ2+1)/ (2+โ2-โ2-1)

2-(โ2+1)/1

2-โ2-1

1-โ2