[Algebra] How do you solve these problems?

Question

In algebra, how do you solve these problems?

Click the Flickr link. The image is easier for seeing the problems.

Questions:

http://www.flickr.com/photos/dawpa2000/2800353772/

2 * (cube root of x) + 9 = 5

(p^-2) - (2p^-1) - 1 = 0

2^(x+3) = 4^(x-1)

I have forgotten what to do. I have been stuck on these problems for a while.

Answer 1

2^(x+3) = 4^(x-1)

4 = 2^2

hence 2^(x+3) = 2^2(x-1)

hence x+3 = 2(x-1)

=> x+3 = 2x -2

=> x = 5

==================================

Answer 2

2 * (cube root of x) + 9 = 5

2*(³√x) + 9 = 5

2*(³√x) = -4

(³√x) = -2

(³√x)³ = (-2)³

x = -8

(p^-2) - (2p^-1) - 1 = 0

1/p² - 2/p - 1 = 0

1/p² - 2p/p² - p²/p² = 0

(1 - 2p - p²)/p² = 0

1 - 2p - p² = 0

-1(p² + 2p - 1) = 0

p² + 2p - 1 = 0

Use quadratic formula:

[-b ± √(b² - 4ac)]/2a where a = 1, b = 2, c = -1

[-2 ± √(2² - 4*1*(-1))]/2*1

[-2 ± √(4 + 4)]/2

[-2 ± √8]/2

[2 ± 2√2]/2

p = 1 ± √2

2^(x+3) = 4^(x-1) ← Note: Get bases the same then set the exponents equal to each other.

2^(x+3) = 2^(2*(x-1))

x + 3 = 2(x - 1)

x + 3 = 2x - 2

-x = -5

x = 5

Check: 2^(x+3) = 4^(x-1)

2^(5+3) = 4^(5-1)

2^8 = 4^4

256 = 256 ← True

x = 5 ← ANSWER

Answer 3

(i) 2 * (cube root of x) + 9 = 5

2*(cube root of x) = -4x^(1/3) = -2

2x^(1/3) = -4

x = (-2)^(3)

   = -8

(ii)(p^-2) - (2p^-1) - 1 = 0

[p^(-2)]( 1 - 2p - p^(2) ) = 0

therefore, p = 0

OR

( 1 - 2p - p^(2) ) = 0

p = (-b(squareroot)(b^(2)-4ac)) / 2a

(ii) 2^(x+3) = 4^(x-1)

(x+3)log2 = (x-1) log4

xlog2 + 3log2 = xlog4 - log4

x(log2 - log4) = -log4 - 3log2

x = (-log4 - 3log2) / (log2 - log4)

x = 5

Answer 4

First one: isolate the ³√x  figure in terms of constants, then cube the constant value thus obtained, to get x.

2³√x + 9 = 5

2 ³√x = −4

³√x = −2

x = −8

Second one: Multiply through by −p², rearrange to standard quadratic equation form, then solve for p.

− 1 + 2p + p² = 0

p² + 2p − 1 = 0

p = 1  ÃƒÂ‚± √2

Third one: Change the 4 to 2².

2³ × 2^x = 2^(2x) × 2^(−2)

Taking binary logarithms...

3 + x = 2x − 2

3 = x − 2

5 = x