**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