Solving the Equation: 2^k - 12^k 2^k1 - 1

Solving the Equation: 2^k - 12^k 2^k1 - 1

Exponential equations can often appear daunting, but with the right techniques, they can be solved systematically. Let's explore how to solve the equation 2^k - 12^k 2^k1 - 1 step-by-step.

Step-by-Step Guide

First, let's reframe the equation for clarity and make it easier to manipulate. The given equation is:

2^k - 12^k 2^k1 - 1

Now, let's rewrite it in a more convenient form. We can express 12^k as 2*2^k. So the equation becomes:

2^k - 2*2^k 2^k1 - 1

Next, we can simplify the left-hand side by factoring out 2^k:2^k(1 - 2) 2^k1 - 1

Further simplify the left-hand side:

-2*2^k 2^k1 - 1

Now, let's isolate the exponential term by adding 2^k1 - 1 to both sides:

2^k1 2*2^k 1

To make it simpler, let's let 2^k x, where x represents the exponential term.

x1 2x 1

Bring all terms to one side to form a quadratic equation:

x1 2x - 1 0

Now, we can factor the quadratic equation:

(x - 1)(x 1) 0

Set each factor to zero and solve for x:

x - 1 0, which gives x 1 x 1 0, which gives x -1

Interpreting the Solutions

We have two potential solutions for x: 1 and -1. However, since 2^k is always positive, we discard the negative solution. Therefore, we have:

x 1

This implies that:

2^k 1

Since 2^k equals 1, and knowing that any exponent of 2 that equals 1 is 0, we equate:

k 0

However, if we look back at the transformed equation, we notice an error in the simplification process. The correct simplified form should be:

x/2 x 2x - 1

Simplify further:

x/2 1, which gives x 2

Recall that x 2^k, so:

2^k 2

This implies that:

k 1

Conclusion

Following the algebraic manipulation and solving the equation step-by-step, we find that the solution to the equation 2^k - 12^k 2^k1 - 1 is:

k 1

This demonstrates the importance of carefully following steps and checking for any potential simplification errors. By understanding these steps, you can solve similar exponential equations and approach more complex algebraic problems with confidence.