How do you solve # (x-2)^(3/4)=8#?

Question

Isaiah Anthony · Accepted Answer

#(x-2)^(3/4)=8#

There's a simple fix, which is:

#((x-2)^(3/4))^(4/3)=8^(4/3)#

#(x-2)^1=(2^3)^(4/3)#

#x-2=2^4=16#

#x=18#,

However, we also have other options:

#(x-2)^(3/4)=8#

#((x-2)^(3/4))^4=8^4# This step is dangerous since we can be adding false solutions.

#(x-2)^3=8^4=4096#

#x^3-6x^2+12x-8=4096#

#x^3-6x^2+12x-4104=0#

We already know that one of the zeros is eighteen:

Thus, we divide the polynomial of degree three by (x-18):

#(x^3-6x^2+12x-4104)/(x-18)=x^2+12x+228#

#x^2+12x+228# can be solved by the quadratic formula:

#x=(-12+-sqrt(12^2-4*228))/2#

#x=(-12+-sqrt(-768))/2#

#x=(-12+-16sqrt(-3))/2#

#x=-6+-8isqrt(3)#

These are false solutions, since #(x-2)^(3/4)=+-8i#. These false solutions were added in the step where it is written :"This step is dangerous since we can be adding false solutions."

Thus, x=18 is the sole solution.

Eli Assouline · Answer

#x=18#

Assuming no complex solution is required

Given:#" "(x-2)^(color(magenta)(3/4))=8# Remember that #(x-2)^(3/4)=root(3)((x-2)^4)# Calculate the cube root on each side to yield #" "(x-2)^(color(magenta)(1/4))= 2# Raise each side by a factor of four to obtain #x-2=2^4 = 16# #x=18#

Genesis Allen · Answer

undefined To solve the equation $ (x-2)^{\frac{3}{4}} = 8 $, you can follow these steps:

1. Raise both sides of the equation to the power of $ \frac{4}{3} $ to cancel out the exponent of $ \frac{3}{4} $.
2. Simplify the expression.
3. Solve for $ x $.

So, the solution is:

$$ x - 2 = 8^{\frac{4}{3}} $$

$$ x - 2 = (2^3)^{\frac{4}{3}} $$

$$ x - 2 = 2^{4} $$

$$ x - 2 = 16 $$

$$ x = 16 + 2 $$

$$ x = 18 $$