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# help

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Let $$x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.$$Find $(x+1)^{48}$.

May 3, 2019

#1
+115
+1

$$x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}. Find (x+1)^{48}$$

$$(x+1)^{48} = 125$$

May 3, 2019
#2
+105654
+1

They can get that from an answer page.

I suspect you got this answer from Wolfram|Alpha or some similar site.

Ideal we want answers to teach but not to facilitate cheating.

So part answers, or solid hints, without the final answer is usually the best way to go.

Encourage the asker to interact with you.  If they want more imput thay should ask for it.

Melody  May 3, 2019
edited by Melody  May 3, 2019
edited by Melody  May 3, 2019
#3
+2

Solve for x:
4/((sqrt(5) + 1) (5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = x

Multiply numerator and denominator of 4/((sqrt(5) + 1) (5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) by sqrt(5) - 1:
(4 (sqrt(5) - 1))/((sqrt(5) + 1) (5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1) (sqrt(5) - 1)) = x

(sqrt(5) + 1) (sqrt(5) - 1) = -1 + 1 sqrt(5) - sqrt(5) + sqrt(5) sqrt(5) = -1 + sqrt(5) - sqrt(5) + 5 = 4:
(4 (sqrt(5) - 1))/(4 (5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = x

(4 (sqrt(5) - 1))/(4 (5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = 4/4×(sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = (sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)):

(sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = x

(sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = x is equivalent to x = (sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)):

x = (sqrt(5) - 1)/((5^(1/4) + 1) (5^(1/8) + 1) (5^(1/16) + 1)) = 0.105823017030235.

[0.105823017030235 +1]^48 =~125

May 3, 2019
#4
+105654
+3

Thanks guest for giving me a hint :)

$$x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}\\ x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}\times \frac{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}\\ x= \frac{4{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(5-1)(\root \of5-1)(\root 4\of5-1)(\root {8}\of5-1)}}\\ x= \frac{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(\root \of5-1)(\root 4\of5-1)(\root {8}\of5-1)}}\times \frac{{(\root \of5+1)(\root 4\of5+1)(\root {8}\of5+1)}}{(\root \of5+1)(\root 4\of5+1)(\root {8}\of5+1)}\\ x= \frac{{(5-1)(\root 2\of5-1)(\root 4\of5-1)(\root {16}\of5-1)}}{{(5-1)(\root 2\of5-1)(\root {4}\of5-1)}}\\ x= \frac{{(\root {16}\of5-1)}}{{1}}\\ x=\root {16}\of5-1\\ x+1=\root {16}\of5\\ (x+1)^{48}=(5^{\frac{1}{16}})^{48}\\ (x+1)^{48}=5^3\\ (x+1)^{48}=125\\$$

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May 4, 2019