Find values of x at which the third item of the expansion (x+x^lgx)^5 is 10^6.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll expand the binomial:

(x+x^lgx)^5= C(5,0)*x^5 + C(5,1)*x^4*(x^lgx) + C(5,2)*x^3*(x^lgx)^2 + ...

As we can see, the 3rd term is C(5,2)*x^3*(x^lgx)^2.

We'll evaluate C(5,2) = 5*(5-1)/2 = 10

C(5,2)*x^3*(x^lgx)^2 = 10*x^3*(x^2*lgx)

We also know that the 3rd term is equal to 10^6.

10*x^3*(x^(2*lgx)) = 10^6

We'll divide by 10:

x^3*(x^(2*lgx)) = 10^5

We notice that the bases of the factors from the left are matching, therefore we'll add its exponents.

x^(3+2lgx) = 10^5

We'll take decimal logarithms both sides:

lg [x^(3+2lgx)] = lg 10^5

We'll apply the power property of logarithms:

(3+2lgx)*lg x = 5*lg 10

But lg 10 = 1

(3+2lgx)*lg x = 5

We'll remove the brackets:

3*lg x + 2*(lg x)^2 - 5 = 0

We'll replace lg x by t:

2t^2 + 3t - 5 = 0

We'll apply quadratic formula:

t1 = [-3+sqrt(9+40)]/4

t1 = (-3+7)/4

t1 = 1

t2 = -10/4

t2 = -5/2

We'll determine x:

lg x = t1 => lg x = 1 => x1 = 10

lg x = t2 => lg x = -5/2 => x2 = 10^(-5/2)

The values of x at which the 3rd term of the given expansion is 10^6, are: {10^(-5/2) , 10}.

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