# If 25^{x – 1} = 5^{2x – 1} – 100, then find the value of x.

Mathematics is not only about numbers but it is about dealing with different calculations involving numbers and variables. This is what basically is known as Algebra. Algebra is defined as the representation of calculations involving mathematical expressions that consist of numbers, operators, and variables. Numbers can be from 0 to 9, operators are the mathematical operators like +, -, ×, ÷, exponents, etc, variables like x, y, z, etc.

### Exponents and Powers

Exponents and powers are the basic operators used in mathematical calculations, exponents are used to simplifying the complex calculations involving multiple self multiplications, self multiplications are basically numbers multiplied by themselves. For example, 7 × 7 × 7 × 7 × 7, can be simply written as 7^{5}. Here, 7 is the base value and 5 is the exponent and the value is 16807. 11 × 11 × 11, can be written as 11^{3}, here, 11 is the base value and 3 is the exponent or power of 11. The value of 11^{3} is 1331.

Exponent is defined as the power given to a number, the number of times it is multiplied by itself. If an expression is written as cx^{y} where c is a constant, c will be the coefficient, x is the base and y is the exponent. If a number say p, is multiplied n times, n will be the exponent of p. It will be written as

**p × p × p × p … n times = p ^{n}**

### Basic rules of Exponents

There are certain basic rules defined for exponents in order to solve the exponential expressions along with the other mathematical operations, for example, if there are the product of two exponents, it can be simplified to make the calculation easier and is known as product rule, let’s look at some of the basic rules of exponents,

- Product Rule ⇢ a
^{n}+ a^{m}= a^{n + m} - Quotient Rule ⇢ a
^{n}/ a^{m}= a^{n – m} - Power Rule ⇢ (a
^{n})^{m}= a^{n × m}or^{m}√a^{n}= a^{n/m} - Negative Exponent Rule ⇢ a
^{-m}= 1/a^{m} - Zero Rule ⇢ a
^{0}= 1 - One Rule ⇢ a
^{1}= a

### If 25^{x – 1} = 5^{2x – 1} – 100, then find the value of x.

**Solution:**

As it is clearly seen, the entire problem statement is asking for a simplification using exponent rules, looking at the expression 25

^{x}^{– 1 }= 5^{2x}^{– 1}– 100, it is observed that 25^{x – 1 }can be written as (5^{2})^{x – 1},(5

^{2})^{x – 1}= 5^{2x – 1 }– 100Power Rule ⇢ (a

^{n})^{m}= a^{n × m}5

^{(2x – 2)}= 5^{(2x – 1)}– 1005

^{(2x-1)}– 5^{(2x-2)}= 1005

^{(2x – 2)}(5 – 1) = 1005

^{(2x – 2)}= 255

^{(2x – 2)}= 5^{2}2x – 2 = 2

2x = 4

x = 2

Therefore, the value of x obtained is 2.

### Similar Problems

**Question 1: Simplify 10(y ^{1})^{5}**

**Solution:**

It is observed that 1 is the exponent of y and 5 is the exponent of y1, and 10 is constant, using the power rule of exponents, it can be written as,

Power Rule ⇢ (a

^{n})^{m}= a^{n × m}10(y

^{1})^{5}= 10y^{(1 × 5)}= 10y

^{5}

**Question 2: What is the value of x, if 36 ^{x – 1} = 6^{2x – 1} – 125?**

**Solution:**

As it is clearly seen, the entire problem statement is asking for a simplification using exponent rules, looking at the expression 36

^{x – 1}= 6^{2x – 1}– 180, it is observed that 36^{x – 1 }can be written as (6^{2})^{x – 1},(6

^{2})^{x – 1}= 6^{2x – 1}– 180Power Rule ⇢ (a

^{n})^{m}= a^{n × m}6

^{(2x – 2)}= 6^{(2x – 1)}– 1806

^{(2x – 2)}(6 – 1) = 366

^{(2x – 2)}= 6^{2}6

^{(2x – 2)}= 6^{2}2x – 2 = 2

2x = 4

x = 2

Therefore, the value of x obtained is 2.

**Question 3: Simplify 30(y ^{6})^{0}**

**Solution:**

It is observed that 6 is the exponent of y and 0 is the exponent of y6, and 30 is constant, using the power rule of exponents, it can be written as,

Power Rule ⇢ (a

^{n})^{m}= a^{n × m}30(y

^{6})^{0}= 30(y^{6 × 0})Applying Zero Rule ⇢ a

^{0}= 130 × 1 = 30