indices qn.?

Question

given that (square root 3)^7 + (square root 3)^5 + (square root 3)^3 + 42(square root 3) may be expressed as 3^k, find the value of k.

ans: 4.5

Answer 1

(√3)^7 = 46∙765 3718
(√3)^5 = 15∙588 457 27
(√3)^3 = 5∙196 152 423
42(√3) = 72∙746 133 92
Now add them together.
(√3)^7 + (√3)^5 + (√3)^3 + 42(√3) = 3^k
140∙296 115 4 = 3^k
Log(140∙296 115 4) = k Log 3
2∙147 0465 6146 = k(0∙477 121 254)
k = 2∙147 0465 6146/ 0∙477 121 254
k = 4∙5

You can also take this approach:

(√3)^7 + (√3)^5 + (√3)^3 + 42(√3) = 3^k
3(√3)^5 + (√3)^5 + (√3)^3 + 42(√3) = 3^k
9(√3)^3 + 3(√3)^3 + (√3)^3 + 42(√3) = 3^k
3*9√3 + 3*3√3 + 3√3 + 42√3 = 3^k
27√3 + 9√3 + 3√3 + 42√3 = 3^k
(√3)(27 + 9 + 3 + 42) = 3^k
(√3)(81) = 3^k
(√3)(3*3*3*3) = 3^k
(√3)(3^4) = 3^k
(3^½)(3^4) = 3^k
3^4∙5 = 3^k
→ k = 4∙5

Answer 2

Call your terms a+b+c+d. Now rewrite these terms
a = 3x3x3xsqrt(3)
b = 3x3xsqrt(3)
c = 3xsqrt(3)
d= 3xsqrt(3)x14
Factor out 3xsqrt(3)= A to get A(9+3+1+14)=27A
But 27= 3^3 and A=3^(3/2), so k= 3+3/2=4-1/2