Question

Solve for \[x\].
\[{4^{x - 1}} \times {\left( {0.5} \right)^{3 - 2x}} = {\left( {\dfrac{1}{8}} \right)^{ - x}}\].

Answer 1

Hint: First we have to know what surds and indices. Mention their laws. Then using laws of surds and indices evaluate or solve for \[x\] from the given equation. First convert each term as a power of \[2\]. After that, express the equation in the form of a power of \[2\]. Equating the powers of \[2\] from the both sides of the obtained equation. We get the value of \[x\].

Complete step by step solution:
Let \[a\] be a rational number and \[n\] be a positive integer such that\[{a^{\dfrac{1}{n}}} = \sqrt[n]{a}\] is irrational. Then, \[\sqrt[n]{a}\] is called \[a\] surd of order \[n\].
Laws of surds: Suppose \[a\]and \[b\]are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds:
A) \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\]
B) \[\sqrt[n]{{ab}} = {a^{\dfrac{1}{n}}} \times {b^{\dfrac{1}{n}}}\]
C) \[\sqrt[n]{{\dfrac{a}{b}}} = \dfrac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}}\]
D) \[\left( {\sqrt[n]{a}} \right) = a\]
E) \[\sqrt[m]{{\left( {\sqrt[n]{a}} \right)}} = \sqrt[{mn}]{a}\]
F) \[{\left( {\sqrt[m]{a}} \right)^n} = \sqrt[m]{{\left( {{a^n}} \right)}}\]
Given \[{4^{x - 1}} \times {\left( {0.5} \right)^{3 - 2x}} = {\left( {\dfrac{1}{8}} \right)^{ - x}}\]----(1)
Since \[4 = {2^2}\] then \[{4^{x - 1}} = {2^{2\left( {x - 1} \right)}} = {2^{2x - 2}}\], and also \[0.5 = \dfrac{5}{{10}} = \dfrac{1}{2}\] then \[{\left( {0.5} \right)^{3 - 2x}} = {\left( {\dfrac{1}{2}} \right)^{3 - 2x}} = \dfrac{1}{{{2^{3 - 2x}}}} = {2^{2x - 3}}\].
We also know that \[\dfrac{1}{8} = \dfrac{1}{{{2^3}}}\] then \[{\left( {\dfrac{1}{8}} \right)^{ - x}} = {\left( {\dfrac{1}{{{2^3}}}} \right)^{ - x}} = \dfrac{1}{{{2^{ - 3x}}}} = {2^{3x}}\].
Then the equation (1) can be rewrite it as
\[{2^{2x - 2}} \times {2^{2x - 3}} = {2^{3x}}\]
Then simplifying the above equation by using laws of indices, we get
\[{2^{2x - 2 + 2x - 3}} = {2^{3x}}\]
\[ \Rightarrow {2^{4x - 5}} = {2^{3x}}\]----(2)
Equating the powers of \[2\] in the equation (2), we get
\[4x - 5 = 3x\]
Solving the above equation, we get
\[x = 5\].
Therefore, the value of $x$ is $5$.

Note:
Laws of indices: Suppose \[a\]and \[b\]are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds:
(i) \[{a^m} \times {a^n} = {a^{m + n}}\]
(ii) \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
(iii) \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
(iv) \[{\left( {ab} \right)^n} = {a^n}{b^n}\]
(v) \[{\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}}\]
(vi) \[{a^0} = 1\]