Substitute the values for x and y within the equation: 5(4)² - 3(5)¹. Proceed according to proper order of operations: 5(16) – 3(5). Therefore: 80-15= 65.

In order to set the exponents equal to each other and solve for t, there must be the same number raised to those exponents.

64 = (√2)ⁿ?

(√2)² = 2 and 2⁶ = 64, so ((√2)²)⁶= (√2)^2*6 = (√2)¹².

Thus, we now have (√2)^12(t+1) = (√2)^{10t + 4}.

12(t+1) = 10t + 4

12t + 12 = 10t + 4

2t + 12 = 4

2t = –8

t = –4

You can simplify it into (x³)(x³) = x⁶

To tackle this problem we must understand the concept of exponents in fractions and how to cancel and move them.

To move any variable or number from the numerator to the denominator or vice versa, you must negate the exponent. i.e.  in the numerator would become  in the denominator. These two expression are equivalent. You should strive to make all exponents positive initially before applying the next rule to simplify.

Cancelling variables with a similar base is an easy way to simplify. Add or subtract the exponents depending on their relationship in a fraction.

Ex.  or 1.

Ex.  . --> this can be more easily understood if you break down the .  which then can be moved around to form, . After the  cancels to form 1, we have  or just . This can be applied for all numerical or abstract values of exponents for a given variable, such as ,  or .

Knowing these rules, we can tackle the problem.

To begin we will pick a variable to start with, thereby breaking down the problem into three smaller chunks. First we will start with the variable . . Because the numerator has a negative exponent, we will move it down to the denominator: . This simplifies to  as multiplying any common variables with exponents is found by addition of the exponents atop the original variable. The  variable part of this problem is .

We move to the  section of the problem: . This is similar to our  above, instead with larger numerical exponents. . The  section cancels, leaving us with  or .

Now to the  section. We simply have  on top. Applying the first rule above, we just move it to the denominator with the switching of the sign. Our result is .

Combining all the sections together we have .

More beautifully written it looks like .

We are given an expression with a power to a power to a power. Using rules of exponents, we take the exponents and multiply each of them together. 

The first step is to distribute the squared on the second term. (2a³)² becomes 4a⁶ by multiplying the exponents (power raised to a power exponent rule) and squaring the 2.  Then, combining like terms (i.e. combining coefficients, a's and b's) we obtain 12a⁸b⁵.

This requires us to remember the rules for multiplying and adding exponent variables.

(3y²)² can be re-written as (3)² x (y²)² which yields 9y⁴

(4y)³ can be re-written as (4)³ x (y)³ which yields 64y³

adding the two yields

9y⁴ +64y³

The answer is (x¹⁰)/2. When an exponent is raised to another power, you multiply the exponents.

 is simplified by multiplying the exponent outside of the expression , with each number and variable inside the expression: 

This gives you  

When variables with exponents are multiplied, you add their respective exponents together, so 

Altogether, the expression is simplified to