Question 204·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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If $2^{x} + 2^{-x} = 10$ , what is the value of $4^{x} + 4^{-x}$ ?

Your answer

For problems where you are given an expression like $a^x + a^{-x}$ and asked for another expression involving the same base and exponent, look for algebraic relationships instead of solving for $x$ . First, rewrite all powers in terms of a single base (for example, rewrite $4^x$ as $(2^x)^2$ ). Then use identities such as $(y + 1/y)^2 = y^2 + 2 + 1/y^2$ to connect what you are given to what you need. This lets you plug in the known value and do straightforward arithmetic, saving time and avoiding messy logarithms or solving for $x$ directly.

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Connect $4^x$ with $2^x$

Rewrite $4$ as a power of $2$ . How can you express $4^x$ and $4^{-x}$ using $2^x$ and $2^{-x}$ ?

Express the target in terms of $(2^x)^2$ and $(2^{-x})^2$

Once $4^x$ and $4^{-x}$ are rewritten, you should get an expression involving $(2^x)^2$ and $(2^{-x})^2$ . How is this related to $2^x + 2^{-x}$ ?

Use the square of a binomial

Consider $(2^x + 2^{-x})^2$ . Expand it using $(a+b)^2 = a^2 + 2ab + b^2$ and simplify the middle term. How can you then isolate $(2^x)^2 + (2^{-x})^2$ ?

Substitute the given value

After you express $(2^x)^2 + (2^{-x})^2$ in terms of $2^x + 2^{-x}$ , replace $2^x + 2^{-x}$ with 10 and simplify.

Desmos Guide

Use the relationship you found to evaluate numerically

After you have written $4^x + 4^{-x}$ as $(2^x + 2^{-x})^2 - 2$ , type (10)^2 - 2 into a Desmos expression line. The value that Desmos shows is the value of $4^x + 4^{-x}$ ; compare it with your answer.

Step-by-step Explanation

Rewrite powers of 4 in terms of powers of 2

Notice that $4 = 2^2$ , so

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

Therefore,

4^x + 4^{-x} = (2^x)^2 + (2^{-x})^2.

So we want $(2^x)^2 + (2^{-x})^2$ .

Relate $(2^x)^2 + (2^{-x})^2$ to $2^x + 2^{-x}$

Consider the square of the sum $2^x + 2^{-x}$ :

(2^x + 2^{-x})^2 = (2^x)^2 + 2(2^x)(2^{-x}) + (2^{-x})^2.

But $(2^x)(2^{-x}) = 2^{x-x} = 2^0 = 1$ , so this becomes

(2^x + 2^{-x})^2 = (2^x)^2 + 2\cdot 1 + (2^{-x})^2 = (2^x)^2 + (2^{-x})^2 + 2.

Rearrange to isolate $(2^x)^2 + (2^{-x})^2$ :

(2^x)^2 + (2^{-x})^2 = (2^x + 2^{-x})^2 - 2.

From Step 1, this is exactly $4^x + 4^{-x}$ .

Substitute the given value and simplify

We are told that $2^x + 2^{-x} = 10$ , so plug this into the expression from Step 2:

4^x + 4^{-x} = (2^x + 2^{-x})^2 - 2 = 10^2 - 2.

Compute:

10^2 - 2 = 100 - 2 = 98.

So the value of $4^x + 4^{-x}$ is 98.

Question 204·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation