If the third term in the binomial expansion of (1 + x^(log2 x))^5 equals 2560, then a possible value

Q: Why is $r=2$ for the third term? I thought it should be 3?

Ah, the classic off-by-one error! Remember, the general term is $T_{r+1}$. So, for the first term, $r=0$. For the second, $r=1$. Therefore, for the third term, $r$ must be $2$. Always keep this index shift in mind!

Q: Why can't I just take the square root of $x^{2 \log_2 x} = 256$ directly?

You could, but you'd be left with $x^{\log_2 x} = 16$. While that's a valid step, you'd still have the variable in the exponent. Using logarithms is the most robust way to linearize the problem and avoid missing the second solution.

Q: Why do we get two values for $x$? Is one of them extraneous?

Great question! In this specific problem, both $x=4$ and $x=1/4$ are valid because the base of the logarithm is positive and the expression is well-defined. Always check the domain of your functions, but here, both roots survive the test!

Question

If the third term in the binomial expansion of $(1 + x^{\log_2 x})^5$ equals 2560, then a possible value of x is :

Options:

(A) 2\sqrt{2}

(B) $\frac{1}{8}$

(C) 4\sqrt{2}

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the binomial expansion of $(1 + x^{\log_2 x})^5$. We are given that the third term, $T_3$, is equal to $2560$.

To solve this, we utilize the general term formula for a binomial expansion:
$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

In this specific case, we identify the parameters as $n = 5$, $a = 1$, and $b = x^{\log_2 x}$.

### The Binomial Foundation

To find the third term $T_3$, we set $r+1 = 3$, which implies $r = 2$. Substituting these values into our general formula, we obtain:

$$T_3 = \binom{5}{2} (1)^{5-2} (x^{\log_2 x})^2 = 2560$$

Since $\binom{5}{2} = \frac{5 	imes 4}{2 	imes 1} = 10$ and $1^3 = 1$, the equation simplifies to:

$$10 \cdot (x^{\log_2 x})^2 = 2560$$

Dividing both sides by $10$, we arrive at:

$$(x^{\log_2 x})^2 = 256$$

### The Logarithmic Twist

Applying the laws of exponents, specifically $(a^m)^n = a^{mn}$, we rewrite the expression as:

$$x^{2 \log_2 x} = 256$$

To solve for $x$, we take the logarithm with base $2$ on both sides of the equation:

$$\log_2(x^{2 \log_2 x}) = \log_2(256)$$

Using the power rule $\log_a(m^n) = n \log_a m$, the expression becomes:

$$(2 \log_2 x) \cdot (\log_2 x) = \log_2(2^8)$$

### Solving the Quadratic

Simplifying the right side, we know that $\log_2(2^8) = 8$. This yields the following quadratic form:

$$2 (\log_2 x)^2 = 8$$

Dividing by $2$, we get:

$$(\log_2 x)^2 = 4$$

Taking the square root of both sides, we find two possible cases for $\log_2 x$:

$$\log_2 x = 2 \quad 	ext{or} \quad \log_2 x = -2$$

### Final Calculation

Converting these logarithmic equations back into exponential form, we solve for $x$:

For $\log_2 x = 2$, we have $x = 2^2 = 4$.

For $\log_2 x = -2$, we have $x = 2^{-2} = \frac{1}{4}$.

Thus, the possible values for $x$ are **$4$** and **$\frac{1}{4}$**.

Animated Solution for Mathematics - Binomial Theorem: If the third term in the binomial expansion of $(1 + x^{l o g_{2} x})^{5}$ equals 2560, then a possible value of x is :

Select Answer:

Visualized Solution

Identify the Binomial Expression

The General Term Formula

Substitute Values into Formula

Simplify the Combinatorics

Isolate the Variable Term

Apply Logarithm to Both Sides

Use Logarithm Properties

Solve for $(lo g_{2} x)^{2}$

Find $lo g_{2} x$

Calculate Final Values of x

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Binomial Foundation

The Logarithmic Twist

Solving the Quadratic

Final Calculation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the third term in the binomial expansion of (1+xlog2​x)5 equals 2560, then a possible value of x is :