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Simplifying Exponents, and Completing the Square


iheartmath

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So I've had a week off from school because of the snow and I finally have to take my IB math midterm tomorrow. I found a few problems in my book that I need someone to explain to me. Your help is appreciated!!!

1. Find the exact value of x for which 4^x * 5^(4x+3) = 10^(2x+3)

2. Solve by completing the square: 4x^2 + 4x = 5

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I don't have time to write and scan my working so can I just type here? Btw I answered your previous questions, did you read my answer?

Qn1)

4x * 54x+3 = 102x+3

22x * 54x * 53 = 22x * 23 * 52x * 53

22x * 54x * 53 = 22x * 23 * 52x * 53

52x = 23

52x = 8

2x = log5 (8)

x = log5 (8) / 2

Calculate by yourself!

Qn2)

4x2 + 4x = 5

0 = 4x2 + 4x - 5

0 = 4 * (x2 + x - 5/4)

0 = 4 * (x2 + x + (1/2)2 - 5/4 - (1/2)2)

0 = 4 * [ (x + 1/2)2 - 6/4 ]

0 = 4(x + 1/2)2 - 6

DONE!

Please check my working btw.. I just woke up and am still sleepy so idk if my Math is correct

Good luck! :gluck:

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  • 3 months later...

Sure, sorry I was at school when you posted that so I think I am too late to reply now :sadnod:

(6n + 12n) / (1 + 2n)

= (2n*3n + 22n*3n) / (1 + 2n)

= (2n*3n (1 + 2n)) / (1 + 2n)

= 2n*3n

= 6n

:gluck:

I'm pretty sure 2n*3n = 62n because you add the powers

So my solution was:

(6n + 12n) / (1 + 2n)

= [2n x 3 + 2n x 6] / ( 1 + 2n)

= [2n ( 6 + 3 )]/ (1 + 2n)

= [2n (9)]/ (1 + 2n)

= 9 / 1

= 9

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