# Thread: [SOLVED] Solve for X

1. ## [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

2. Originally Posted by mathman619
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
If that is
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
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.

3. Originally Posted by mathman619
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
Hello,

only in case you mean:
$\displaystyle 2^{\frac{2}{3} x+1}-3\cdot 2^{\frac{2}{3}x} -20 = 0$

then do the substitution as ticbol has told you. You'll get:
$\displaystyle y=2^{\frac{1}{3}x}$ . So you have:
$\displaystyle 2 \cdot y^2 - 3 \cdot y -20=0$

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

Now you've to re-substitute
$\displaystyle 4=2^{\frac{1}{3}x}\ \Leftrightarrow \ 2^2=2^{\frac{1}{3}x}\ \Rightarrow \ x=6$

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