Solve This Problem & Be Hired as Consultant - Jobs/Vacancies (2)

Solve This Problem & Be Hired as Consultant - Jobs/Vacancies (2) - Nairaland

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Re: Solve This Problem & Be Hired as Consultant by ceejayluv(m): 5:05pm On Oct 06, 2015*

got the answer, lemme type it....

*modified.... answer didn't add up... mission aborted, goodnight

2 Likes

Re: Solve This Problem & Be Hired as Consultant by tee234: 5:16pm On Oct 06, 2015

2xlog2=xlog1 + log8
X(2log2 - log 1)=log8
X0.602=0.903
X=1.5

1 Like

Re: Solve This Problem & Be Hired as Consultant by Adehorze(m): 5:19pm On Oct 06, 2015

2^2x=8x
Taking log of both sides,
Log2^2x=Log8x
2xLog2=Log8+Logx
2xLog2=Log2^3+Logx
2xLog2=3Log2+Logx
2xLog2-3Log2=Logx
(2x-3)Log2=Logx
Equating d log of both sides,
>>
(2x-3)2=x
4x-6=x
4x-x=6
3x=6
Dividing both sides by 3
3x/3=6/3
x=2

8 Likes 2 Shares

Re: Solve This Problem & Be Hired as Consultant by jeff2010(m): 5:19pm On Oct 06, 2015

2^2x=8x 2^2x=2^2.2x Take log Log 2^2x=log 2^2.2x Log2^2x=log2^2 + log2x 2xlog2=2log2 + log2x (2x-2)log2= log2x 2x-2=x 2x-x=2 X=2 Did I get it huh

Re: Solve This Problem & Be Hired as Consultant by jeff2010(m): 5:19pm On Oct 06, 2015

...

Re: Solve This Problem & Be Hired as Consultant by agabaI23(m): 5:19pm On Oct 06, 2015

This is guess work but I think it is logical

2^2x=8x

2^2x=(2^3)x (2 raised to 2x is the same as 2 raised to power 2+x. Solving 2+x=3; x=1)

If x =1
Then
2^2+1= (2^3)x1

so 2^3=2^3
8=8

I am not sure, it looks to me like carpentered stuff grin

Re: Solve This Problem & Be Hired as Consultant by Mekyno(m): 5:24pm On Oct 06, 2015

Adehorze:
2^2x=8x Taking log of both sides, Log2^2x=Log8x 2xLog2=Log8+Logx 2xLog2=Log2^3+Logx 2xLog2=3Log2+Logx 2xLog2-3Log2=Logx (2x-3)Log2=Logx Equating d log of both sides, >> (2x-3)2=x 4x-6=x 4x-x=6 3x=6 Dividing both sides by 3 3x/3=6/3 x=2

gr8.....tnk God n ur parents 4 me

Re: Solve This Problem & Be Hired as Consultant by Mekyno(m): 5:29pm On Oct 06, 2015

Neyozak:
2raised to power 2x = 8x

make sure u reward d winner ooh. Let it nt be a ploy to make guys wake their maths spirit without reward oooh.

Re: Solve This Problem & Be Hired as Consultant by chiv: 5:37pm On Oct 06, 2015

Just got the answer
2^2x=8x
2^2(x)=8x
4^x=8x........eqn(1)
2^2x=8x
2^2x=(2^3)x
Divide both sides by 2^3
x=2^(2x-3)........eqn(2)
Substitute 2^2(2x-3) for x in eqn(1)
4^(2^(2x-3))=8(2^(2x-3))
16^(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^2x)/2^3
2^4(2x-3)=2^2x
Equate the powers of 2 on both sides
4(2x-3)=2x
8x-12=2x
-12=2x-8x
-12=-6x
x=2

2 Likes

Re: Solve This Problem & Be Hired as Consultant by Nobody: 7:15pm On Oct 06, 2015

oya op come see am
2^2x=8x. take log
log2^2x=log8x
2xlog2=log8+logx
2x=[log8+logx]/log2
(after division by log 2, the base changes from 10 to 2)
2x-log=log8. (all in base 2 henceforth)
since 2x=2xlog2 or xlog4
xlog4-logx=log8
xlog4-logx=log4+log2

xlog4-log4=log2+logx

log4(x-1)=log2x

4(x-1)=2x
4x-2x=4

x=2

Re: Solve This Problem & Be Hired as Consultant by ZACHIE: 8:29pm On Oct 06, 2015

demolinka:
oya op come see am

2^2x=8x. take log

log2^2x=log8x

2xlog2=log8+logx

2x=[log8+logx]/log2

(after division by log 2, the base changes from 10 to 2)

2x-log=log8. (all in base 2 henceforth)

since 2x=2xlog2 or xlog4

xlog4-logx=log8

xlog4-logx=log4+log2

xlog4-log4=log2+logx

log4(x-1)=log2x

4(x-1)=2x

4x-2x=4

x=2

Yes.
Final amswer++++
OBATALA approved it

Re: Solve This Problem & Be Hired as Consultant by darium: 8:40pm On Oct 06, 2015

2~2x = 8x
Take log of both sides
Log 2~2x =log 8x
2xlog2 = log8 + log x
2xlog2 = 3log2 + log x
2xlog2 - 3log2 = log x
Log2(2x-3) = log x
Take anti-log of both sides
2(2x-3) = x
4x - 6 = x
4x - x = 6
3x = 6
X = 2

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Re: Solve This Problem & Be Hired as Consultant by Nobody: 8:53pm On Oct 06, 2015

Olodos in the house nah 3/2 be the answer SS1 work nah u dey Bang

Re: Solve This Problem & Be Hired as Consultant by Asiwaju9ja(m): 8:59pm On Oct 06, 2015

dejt4u:
2^2x=8x,

2^2x=2^3(x),

(2^2x) / (2^3) = x,

2^(2x-3)=x,

2^(2x-3)/4^(2x-3)=x/4^(2x-3),

(1/2)^(2x-3)=x*(1/4)^(2x-3),

(0.5)^(2x-3)=x*(0.25)^(2x-3),

(1-0.5)^(2x-3)=x*(1-0.75)^(2x-3),

by binomial approx.

1-(2x-3)*0.5...=

x*[1-(2x-3)*0.75+...]

re-arrangin gives;

1-x+3/2=x-3x/4*(2x-3) = 4-4x+6=4x-6x^2+9x

or

6x^2-17x+6=0,

x=2.4 or - 0.42

Reference to laplacian of nairaland maths clinic

Una still dey find "X" ? Even Dem Azikiwe still try find am for their time.

Meanwhile, d "x" dey after after 2 and 8 for the equation abi una no see am?

Re: Solve This Problem & Be Hired as Consultant by Mstcambridge: 9:23pm On Oct 06, 2015

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Re: Solve This Problem & Be Hired as Consultant by smeag0l(m): 9:37pm On Oct 06, 2015

I can solve it only if you tell me the kind of job and the pay. I wouldn't want to waste my time on some silly simple maths that wouldn't yield anything to my bank account

Neyozak:
But u tried sha! But not convincing. Idea my guy. 50% 4 u.

Re: Solve This Problem & Be Hired as Consultant by singlefade25(f): 9:51pm On Oct 06, 2015

EleDeVee

Re: Solve This Problem & Be Hired as Consultant by AK461(m): 10:54pm On Oct 06, 2015

Answer: x = 2

2^2x=8x

2^2(2)=8(2)

2^4=8*2

16=16
__________________
QED

1 Like

Re: Solve This Problem & Be Hired as Consultant by 4stylz: 5:48am On Oct 07, 2015

If everyone has to go through this stress /wahala to get hired, do you think anyone will ever get a job?

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 7:17am On Oct 07, 2015*

Adehorze:
2^2x=8x
Taking log of both sides,
Log2^2x=Log8x
2xLog2=Log8+Logx
2xLog2=Log2^3+Logx
2xLog2=3Log2+Logx
2xLog2-3Log2=Logx
(2x-3)Log2=Logx
U CAN'T EQUATE D LOGS HERE TO HAVE D NEXT STEP.
Equating d log of both sides,
>>
(2x-3)2=x
4x-6=x
4x-x=6
3x=6
Dividing both sides by 3
3x/3=6/3
x=2

After a careful consideration, an error discovered. However u tried.

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 7:26am On Oct 07, 2015*

chiv:
Just got the answer
2^2x=8x
2^2(x)=8x
4^x=8x........eqn(1)
2^2x=8x
2^2x=(2^3)x
Divide both sides by 2^3
x=2^(2x-3)........eqn(2)
Substitute 2^2(2x-3) for x in eqn(1)
4^(2^(2x-3))=8(2^(2x-3))
16^(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^2x)/2^3
2^4(2x-3)=2^2x
Equate the powers of 2 on both sides
4(2x-3)=2x
8x-12=2x
-12=2x-8x
-12=-6x
x=2

So far so good, this is the best approach. Invitation loading.

1 Like

Re: Solve This Problem & Be Hired as Consultant by Nobody: 7:47am On Oct 07, 2015

Maths.......my one and only enemy

1+1 =11 cry

1 Like

Re: Solve This Problem & Be Hired as Consultant by Godschild2(m): 8:53am On Oct 07, 2015*

Okay, lemme try my own too...

2^2x = 8x
2^2x = 2³x
(divide both sides by 2³)
2^2x/2³ = x
2^2x-3 = x¹
(since any number raised to power of 1 is the number itself, 'x' on the R.H.S above can be any number. So we let x=2)
2^2x-3 = 2¹

Therefore;
2x-3 = 1
2x = 4
x = 4/2
x = 2

Re: Solve This Problem & Be Hired as Consultant by Godschild2(m): 9:22am On Oct 07, 2015

...Another method;

2^2x = 8x
2^2x = 2³x¹
(multiply both sides by 1, since x⁰=1 on the L.H.S)
2^2x.x⁰ = 2³.x¹
(divide both sides by 2x so the exponents will remain)
2x+0 = 3+1
2x = 4
x = 2

Re: Solve This Problem & Be Hired as Consultant by onlynaija: 10:06am On Oct 07, 2015

My Solution

Re: Solve This Problem & Be Hired as Consultant by Nobody: 10:22am On Oct 07, 2015

Via numerical method (The Lambert W function to be precise) There are two solutions to this equation: x=0.154953, x=2.

1 Like

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 10:38am On Oct 07, 2015

Godschild2:
...Another method;
2^2x = 8x 2^2x = 2³x¹ (multiply both sides by 1, since x⁰=1 on the L.H.S) 2^2x.x⁰ = 2³.x¹ (divide both sides by 2x so the exponents will remain) 2x+0 = 3+1 2x = 4 x = 2

U just manipulated, however u tried.

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 10:43am On Oct 07, 2015

Godschild2:
Okay, lemme try my own too...

2^2x = 8x
2^2x = 2³x
(divide both sides by 2³)
2^2x/2³ = x
2^2x-3 = x¹
(since any number raised to power of 1 is the number itself, 'x' on the R.H.S above can be any number. So we let x=2)
2^2x-3 = 2¹

Therefore;
2x-3 = 1
2x = 4
x = 4/2
x = 2

U also tried but not too convincing.

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 10:44am On Oct 07, 2015

Karmanaut:
Via numerical method (The Lambert W function to be precise)
There are two solutions to this equation:
x=0.154953, x=2.

Okay I hear

Re: Solve This Problem & Be Hired as Consultant by Neyozak(op): 10:47am On Oct 07, 2015

Thread Closed. I sincerely wanna thank all that contributed. And I guess someone has learn one to things.
@ Chiv pls send ur whatup no to okunadeniyi@gmail.com
Great mathematicians! I thank u all once again.

Re: Solve This Problem & Be Hired as Consultant by beau49: 10:50am On Oct 07, 2015

s.inglefade25:
EleDeeVee

Re: Solve This Problem & Be Hired as Consultant by prideofscience: 11:10am On Oct 07, 2015

Tanking to log to base 10 of both sodes
log100^2x = log10^8x

given the same base

log10^2*2x =log10^8x
rearranging the terms
Log10^2*2x - Log10^8x =0
2+2x / 8x =0

2 + 2x =8x X0 ........ 2x = -2, div tru by 2

x = -2/2
ans x = -1

1 2 3 4 5 6 Reply

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