Question

If \[{125^x} = \dfrac{{25}}{{{5^x}}} \] , then find \[2x \]

Answer 1

Hint: First we have to know the \[\log \] basic formulas. Then taking \[\log \] on both sides of the given equation, using the \[\log \] formulas reduced it as much as possible. Using surds and indices laws find the value of \[x \] . Then find the value of \[2x \] . There is another method, first convert each term in the given equation as a power of 5 and using the laws of indices and surds we get the value of x, then find the value of \[2x \] .

Complete step-by-step solution:
Suppose \[p \] and \[q \] are any two non-zero positive real numbers the following formulas holds:
 \[{\log _n}\left( {pq} \right) = {\log _n}\left( p \right) + {\log _n}\left( q \right) \] .
 \[{\log _n}\left( {\dfrac{p}{q}} \right) = {\log _n}\left( p \right) - {\log _n}\left( q \right) \] .
 \[{\log _n}\left( {{p^a}} \right) = a{\log _n}\left( p \right) \] .
 \[{\log _p}\left( q \right) = \dfrac{{{{\log }_n}\left( q \right)}}{{{{\log }_n}\left( p \right)}} \] .
Given \[{125^x} = \dfrac{{25}}{{{5^x}}} \] ----(1)
Taking \[\log \] on the both sides of the equation (1), we get
 \[\log \left( {{{125}^x}} \right) = \log \left( {\dfrac{{25}}{{{5^x}}}} \right) \] ----(2)
Using the \[\log \] formulas the equation (2) becomes
 \[x\log \left( {125} \right) = \log \left( {25} \right) - x\log (5) \] ---(3)
Using the surds and indices laws the equation (3) becomes
 \[x\log \left( {{5^3}} \right) = \log \left( {{5^2}} \right) - x\log (5) \] ---(4)
Again, using the \[\log \] formulas the equation (2) becomes
 \[3x\log \left( 5 \right) = 2\log \left( 5 \right) - x\log (5) \]
 \[ \Rightarrow 4x = 2 \]
 \[ \Rightarrow x = \dfrac{1}{2} \] .

Hence, the value \[2x \] is \[1 \] .

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:
 \[{a^m} \times {a^n} = {a^{m + n}} \]
 \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \]
 \[{\left( {{a^m}} \right)^n} = {a^{mn}} \]
 \[{\left( {ab} \right)^n} = {a^n}{b^n} \]
 \[{\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}} \]
 \[{a^0} = 1 \]
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:
 \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}} \]
 \[\sqrt[n]{{ab}} = {a^{\dfrac{1}{n}}} \times {b^{\dfrac{1}{n}}} \]
 \[\sqrt[n]{{\dfrac{a}{b}}} = \dfrac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}} \]
 \[\left( {\sqrt[n]{a}} \right) = a \]