Solve:

(1/81)^t=243^t-2

The book states to get the same base but I don't understand.

Thanks for any help.

the key is to recognize

both 81 and 243 as powers of 3

81 = 3^4 and 243 = 3^5

so (1/81)^t=243^t-2
(3^-4)^t = (3^5)^(t-2)

3^(-4t) = 3^(5t-10)

then -4t = 5t - 10
-9t=-10
t=10/9

To solve the equation (1/81)^t = 243^(t-2), we can follow the steps mentioned in the book to simplify and solve it. The approach involves getting the same base on both sides of the equation.

1. Start by observing the bases on both sides. On the left side, the base is 1/81, and on the right side, the base is 243. These two bases are not the same.

2. We can rewrite 1/81 as (1/3)^4 and rewrite 243 as (3^5). This step is to express both bases using the same prime number (which is 3 in this case).

Therefore, the equation becomes ((1/3)^4)^t = (3^5)^(t-2).

3. Simplify the exponents on both sides. On the left side, (1/3)^4 raised to the power of t is equal to (1/3)^(4t), and on the right side, (3^5) raised to the power of (t-2) is equal to 3^(5t-10).

The equation now becomes (1/3)^(4t) = 3^(5t-10).

4. With both sides of the equation now having the same base (3), we can drop the base and equate the exponents. This step follows the exponent rule that states if two powers with the same base are equal, their exponents must be equal.

Therefore, 4t = 5t - 10.

5. Simplify and solve the equation for t. Subtract 4t from both sides:

4t - 4t = 5t - 10 - 4t,
0 = t - 10.

Add 10 to both sides:

10 = t.

Therefore, the solution to the equation (1/81)^t = 243^(t-2) is t = 10.

I hope this explanation helps you understand the steps involved in solving the equation.