Solve For X

2^(2/3)x+1-3*2^(1/3)x-20=0

^ means to the power of

-3 and -20 are separate from the exponents

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- Mar 10th 2006, 10:25 AMmathman619[SOLVED] Solve for X
Solve For X

2^(2/3)x+1-3*2^(1/3)x-20=0

^ means to the power of

-3 and -20 are separate from the exponents - Mar 10th 2006, 10:48 AMticbolQuote:

Originally Posted by**mathman619**

2^[(2/3)(x+1)] -3*2^[(1/3)x] -20 = 0, ---------(i)

then,

2^[(2/3)x] *2^(2/3) -2^[(1/3)x] *3 -20 = 0 ----(ii)

Let y = 2^[(1/3)x]

Then, (ii) becomes

[2^(2/3)]y^2 -3y -20 = 0

Use the Quadratic Formula,

y = {-(-3) +,-sqrt[(-3)^2 -4(2^(2/3))(-20)]} / [2*2^(2/3)]

y = {3 +,-sqrt(135.9920842)} /(3.174802104)

y = {3 +,-11.6615644}/(3.174802104)

y = 4.618 or -2.728

Umm, 0.618.....

When y = 4.618,

2^[(1/3)x] = 4.618

2^(x/3) = 4.618

Take the common log of both sides,

(x/3)log(2) = log(4.618)

x/3 = log(4.618)/log(2)

x = 3log(4.618)/log(2)

x = 6.6218 --------------***

When y = -2.728,

2^(x/3) = -2.728

log(2^(x/3)) = log(-2.728)

Cannot be. There is no logarithm of negative numbers.

Check x=6.6218 against the original equation,

2^[(2/3)(x+1)] -3*2^[(1/3)x] -20 = 0 ---------(i)

2^[(2/3)(6.6218 +1)] -3*2^[(1/3)(6.6218)] -20 =? 0

2^(5.0812) -3*2^(2.2072667) -20 =? 0

33.853 -13.854 -20 =? 0

-0.001 =? 0

Yes (that -0.001 is due to rounding of the decimals), so, OK.

Therefore, x = 6.6218 -----------answer. - Mar 10th 2006, 11:15 AMearbothQuote:

Originally Posted by**mathman619**

only in case you mean:

then do the substitution as ticbol has told you. You'll get:

. So you have:

Solve for y and you'get: y = 4 or y = -(5/2).

Now you've to re-substitute

The negative value for y is not possible with your equation, because with the positive base of 2 you'll never get a negative result.

So the only possible result is x = 6.

Greetings.

EB