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Solve.






I notice that at the mention of these name, Front page is sure, lemme try it.

Cc: Lalasticlala Fynestboi Richiez

How does d solving affect the price of garri? Jubilating ijebu dance first 2 comment

I will try and Solve it Later!!

Cheap.

Dis is maths? Loool

OP, is that your assignment

C'mon don't be lazy, do your homework now!!!

Acekidc4:
I will try and Solve it Later!!

hmmmmmmm

RickRichards:
Cheap.

cheap?

Solve it then, mtcheew

CountDracula:
Dis is maths? Loool

no! It's Chemistry.

Basildvalour:
OP, is that your assignment


C'mon don't be lazy, do your homework now!!!

mtcheeeewwwwwwwww, some people shaa

amhackdee:
no! It's Chemistry.

lol, come see maths

Question 2

2^x + 32{2^(-x)} - 12 = 0

Rearranging gives :::>

2^x + 32(1/2^x) - 12 = 0

Put 2^x = s

Substituting into the equation gives us:::
s + 32/s - 12 = 0
s² + 32 - 12s = 0

:. s² - 12s + 32 = 0
Factoring gives :::

s² - 8s - 4s + 32 = 0
s(s - 8 ) - 4(s - 8 ) = 0
:. s - 8 = 0 or s - 4 = 0

s = 8 or s = 4

Recall that s = 2^x
Putting back gives :::>

2^x = 8 or 2^x = 4
2^x = 2³ or 2^x = 2²

Thus ::::>

x = 3 or x = 2


Question 4

{3ⁿ - 3^(n-1)}/{3³× 3ⁿ - 27 × 3^(n-1)}

Let's first apply the law of indices ::::

3^(n - 1) can be rewritten as ::: 3ⁿ ÷ 3¹
= 3ⁿ ÷ 3

So that the expression becomes :::

{3ⁿ - 3ⁿ ÷ 3}/{3³× 3ⁿ - 27 × 3ⁿ ÷ 3}

Then let's put 3ⁿ = s

Substituting 3ⁿ = s back into the expression will give us :::

{s - (s ÷ 3)}/{3³× s - 27 × s ÷ 3}

Factoring s in both the numerator and denominator, gives ::::

s{1 - (1/3)}/s{3³ - 27/3}
with s cancelling out, leaves us with::::
:. {1 - (1/3)}/{27 - 9)
= (2/3)÷18
= (2/3)(1/18)
= (1/3)(1/9)
= 1/27

Questions 1 and 3 are not clear! I will type out the solutions for 5 and 6 later!

I hope it helps.