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**Hint**: Here the question is in the form of exponents and where its power are variables. We have to determine the value of the variable. So first we simplify the given inequalities and we try to make the base number of both LHS and RHS the same and we are solving for x.

**Complete step by step solution:**

An exponent refers to the number of times a number is multiplied by itself.

Now we consider the given questions and solve them.

1. \[{25^{x - 1}} = {5^{2x - 1}} - 100\]

The number 25 can be written as \[{5^2}\], because when 5 is multiplied twice we get 25. So we have

\[ \Rightarrow {5^{2\left( {x - 1} \right)}} = {5^{2x - 1}} - 100\]

On simplifying the exponent term we have

\[ \Rightarrow {5^{2x - 2}} = {5^{2x - 1}} - 100\]

Now take 100 to LHS and \[{5^{2x - 2}}\] to RHS, so now we have

\[ \Rightarrow 100 = {5^{2x - 1}} - {5^{2x - 2}}\]

Now take \[{5^{2x - 2}}\] as common in RHS.

\[ \Rightarrow 100 = {5^{2x - 2}}(5 - 1)\]

On subtracting 1 from 5 we have

\[ \Rightarrow 4 \times {5^{2x - 2}} = 100\]

On dividing by 4

\[ \Rightarrow {5^{2x - 2}} = 25\]

The number 25 can be written as \[{5^2}\], because when 5 is multiplied twice we get 25.

\[ \Rightarrow {5^{2x - 2}} = {5^2}\]

The base value are same so we can equate the exponent terms

\[ \Rightarrow 2x - 2 = 2\]

Take 2 to RHS

\[ \Rightarrow 2x = 4\]

On dividing by 2

\[ \Rightarrow x = 2\]

2. \[{5^{x - 3}} \times {3^{2x - 8}} = 225\]

By the law of exponents \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\], we rewrite the LHS terms and the number 225 can be written as a product of 25 and 9

\[ \Rightarrow \dfrac{{{5^x}}}{{{5^3}}} \times \dfrac{{{3^{2x}}}}{{{3^8}}} = 25 \times 9\]

Now take \[{5^3}\,\,and\,\,{3^8}\] to RHS

\[ \Rightarrow {5^x} \times {3^{2x}} = 25 \times 9 \times {5^3} \times {3^8}\]

The number 25 and 9 can be written in the terms of exponents

\[ \Rightarrow {5^x} \times {3^{2x}} = {5^2} \times {3^2} \times {5^3} \times {3^8}\]

In RHS the base number are same we can add the exponents

\[ \Rightarrow {5^x} \times {3^{2x}} = {5^5} \times {3^{10}}\]

On equating the exponents of LHS and RHS we get

\[ \Rightarrow x = 5\]

3. \[{\left( {{5^0} + \dfrac{2}{3}} \right)^2} = {\left( {0.6} \right)^{3 - 2x}}\]

By the law of exponent \[{a^0} = 1\], we rewrite the term of LHS and the decimal number can be written in the form of fraction, on considering this the above inequality is written as

\[ \Rightarrow {\left( {1 + \dfrac{2}{3}} \right)^2} = {\left( {\dfrac{6}{{10}}} \right)^{3 - 2x}}\]

On simplifying the above terms we have

\[ \Rightarrow {\left( {\dfrac{5}{3}} \right)^2} = {\left( {\dfrac{3}{5}} \right)^{3 - 2x}}\]

In the RHS we change the sign of the exponent and numerator and denominator value.

\[ \Rightarrow {\left( {\dfrac{5}{3}} \right)^2} = {\left( {\dfrac{5}{3}} \right)^{2x - 3}}\]

On equating the powers or exponents

\[ \Rightarrow 2 = 2x - 3\]

On simplifying we get

\[ \Rightarrow x = \dfrac{5}{2}\]

4. \[{2^3}({5^0} + {3^{2x}}) = 8\dfrac{8}{{27}}\]

In the LHS use the law of exponent \[{a^0} = 1\] and in RHS convert mixed fraction into improper fraction

\[ \Rightarrow {2^3}(1 + {3^{2x}}) = \dfrac{{224}}{{27}}\]

Expand the exponent term in LHS and in RHS write the number 27 in the form of exponent

\[ \Rightarrow 8(1 + {3^{2x}}) = \dfrac{{224}}{{{3^3}}}\]

Divide the above inequality by 8 we get

\[ \Rightarrow (1 + {3^{2x}}) = \dfrac{{28}}{{{3^3}}}\]

Take \[{3^3}\] to LHS

\[ \Rightarrow {3^3}(1 + {3^{2x}}) = 28\]

On simplifying

\[ \Rightarrow {3^3} + {3^{2x + 3}} = 28\]

\[ \Rightarrow 27 + {3^{2x + 3}} = 28\]

On further simplifying we get

\[ \Rightarrow {3^{2x + 3}} = 1\]

By using the law of exponent 1 can be written as \[{3^0}\]

\[ \Rightarrow {3^{2x + 3}} = {3^0}\]

On equating the powers

\[ \Rightarrow 2x + 3 = 0\]

on simplifying we get

\[ \Rightarrow x = - \dfrac{3}{2}\]

Hence we have determine the value of x for the four different problems

**Note**: The number can be written in the form of the exponents and the exponent number can also be written in the form of the usual number. We must know about the law of exponents. Since the given terms are exponential, we need a law of exponents to solve.

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