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Great Nairaland Mathematicians, your humble student is here again! Kindly solve the following questions.

NB: Please endeavour to solve to a layman's understanding (no mathematical shortcuts (lol)). You can use any convenient method, may be by solving it on a paper and send the solution as a picture or solve it online.

Thanks in anticipation, may God Almighty increase you in knowledge.

Solve the following equations:

1) logP - log30 + log10.8 = 2log1.2

2) SOLVED: Thanks to @Karmanaut

3) If P = log_qR ; Q = log_rP ; R = log_pQ

Show that PQR = 1

4) Solve simultaneously:

A) 2log₁₀X - log₁₀Y = 1
3x + 2y = 20

B) 4logx - logy = 1
5logx + 3logy = 14

5) Show that:
(P+Q)² = 4abq
given that 47pq = p² +
q²


6) Given that a² + b² = 7ab

Prove that 2log(a+b / 3) = loga + logb

INDICES

1) 8ⁿ-4^n-1
_______________________________
16x2ⁿ-4X12^3n-3


2) 12^3n-2x27³ⁿx8³ⁿ
__________________________
2²ⁿx36ⁿx12ⁿ

2.)
log(2x+3) + log(5x-6) =log(10x^2-3x-5)
Recall, from the law of logarithms that log A + log B = Log(A*B)
Therefore:
Log{(2x+3)(5x-6)}= Log(10x^2-3x-5)
Log cancels Log so you are left with:
(2x+3)(5x-6) = 10x^2 - 3x - 5
Expanding the left side you have:
10x^2 -12x + 15x - 18 = 10^2 -3x - 5
(10 x^2 cancels out)
3x - 18 = -3x - 5
6x = -5 + 18
6x = 13
X
Therefore, X = 13/6




Typing this will be long and tedious, and I don't have access to a pen or paper.
When. I'm less busy I might solve it and upload.
Cheers.

The time whey teacher dey teach u for class where your mind dey?abi for text book them write (goto nairaland to get the answers)??* anyway here is the answers


1). A is equal to 5b
2) 5.6
3) 0
4) 8x
5)6 allover 89
6) Go find am for another place I don try

Question 3 is a trivial proof. You can do that one by yourself.

Karmanaut:
2.)
log(2x+3) + log(5x-6) =log(10x^2-3x-5)
Recall, from the law of logarithms that log A + log B = Log(A*B)
Therefore:
Log{(2x+3)(5x-6)}= Log(10x^2-3x-5)
Log cancels Log so you are left with:
(2x+3)(5x-6) = 10x^2 - 3x - 5
Expanding the left side you have:
10x^2 -12x + 15x - 18 = 10^2 -3x - 5
(10 x^2 cancels out)
3x - 18 = -3x - 5
6x = -5 + 18
6x = 13
X
Therefore, X = 13/6




Typing this will be long and tedious, and I don't have access to a pen or paper.
When. I'm less busy I might solve it and upload.
Cheers.

Thank you bros. You are not like some internationally recognized idiots who use any slight opportunity to insult people.
Kindly help solve others as you said, it will go a long way in helping me mathematically especially on this topic.

More blessing bros

peterovanski:
The time whey teacher dey teach u for class where your mind dey?abi for text book them write (goto nairaland to get the answers)??* anyway here is the answers


1). A is equal to 5b
2) 5.6
3) 0
4) 8x
5)6 allover 89
6) Go find am for another place I don try

Respect yourself

Your contributions will be highly appreciated

menzo4u:
Thank you bros. You are not like some internationally recognized idiots who use any slight opportunity to insult people.
Kindly help solve others as you said, it will go a long way in helping me mathematically especially on this topic.

More blessing bros

Before I continue, is this a test, assignment or something?

datz question 1

Question 4:
2logX - log Y = 1 ..... equation (i)
3x + 2Y = 20 .... equation (ii)

Starting from equ 1:
From law of logarithm, (nlogm = logm^n)
Logx^2 - log Y = Log10
(Law of logarithms: LogA-LogB = Log(A/B))
Log (x^2/y) =Log 10
Log cancels out.
X^2/Y = 10
Y = x^2/10

Substitute this for Y in equation 2. (You can make X the subject of the formula and substitute, and still get the same answer-- Maths is wonderful)
3x + 2(x^2/10) = 20
3X + X^2/5 = 20
Taking LCM across equality sign:

15x + X^2 = 100
(Solve the quadratic equation, you'll get two values of X)
Substitute each value of X in Equation 2, and you'll get two corresponding values of Y.)
One pair of values may be extraneous; ie does not satisfy the equation.
Cheers.

solution

continuation

Karmanaut:
Before I continue, is this a test, assignment or something?

it's just for practice, in preparation for an exam....it's from the lecturer's textbook and he said we should practice the questions that he is likely to set it from there or if we are able to solve the questions, we will be able to solve his exam questions.

menzo4u:
it's just for practice, in preparation for an exam....it's from the lecturer's textbook and he said we should practice the questions that he is likely to set it from there or if we are able to solve the questions, we will be able to solve his exam questions.

OK.
Good luck.

agnespommy:
datz question 1

You are blessed bros. Please, I am not just good at Mathematics but once I get answers like this, I can always cope.....op u won't mind helping some other time?

Karmanaut:
OK.
Good luck.

Thanks bros...Maths is just one area that I am quiet weak unlike other courses, hope you won't mind helping someother tilme?

(a+b)^2- 2ab=7ab
=>(a+b)^2=9ab
taking square root
=>(a+b)=3(ab)^1/2
taking log
log(a+b)=l0g3+ 1/2(log(ab))
=>l0g(a+b)-log3=1/2[log a+ log b]
=>log[1/3(a+b)]=1/2[log a+ log b]

menzo4u:
Thanks bros...Maths is just one area that I am quiet weak unlike other courses, hope you won't mind helping someother tilme?

If Im free, no problem; I love maths!

Agnespommy and Nobody...please could you gentlemen give me your whatsapp numbers...I want to add you two please

Ok