Question

The value of \[x\] satisfying the equation \[{\log _{17}}{\log _5}\left( {5\sqrt x - \sqrt {25x - 4} } \right) = 0\] is …………… Find \[50 \times \sqrt x \]?

Answer 1

Hint: Take the logarithmic terms to right-hand side twice to obtain an equation in terms of \[x\] by using the formula \[{\log _a}x = y \Leftrightarrow x = {a^y}\]. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:
Given \[{\log _{17}}{\log _5}\left( {5\sqrt x - \sqrt {25x - 4} } \right) = 0\]
Taking \[{\log _{17}}\]to the right-hand side, we have
\[{\log _5}\left( {5\sqrt x - \sqrt {25x - 4} } \right) = {17^0}{\text{ }}\left[ {\because {{\log }_a}x = y \Leftrightarrow x = {a^y}} \right] \\
  {\log _5}\left( {5\sqrt x - \sqrt {25x - 4} } \right) = 1 \]
Taking \[{\log _5}\] to the right-hand side, we have
\[ 5\sqrt x - \sqrt {25x - 4} = {5^1}{\text{ }}\left[ {\because {{\log }_a}x = y \Leftrightarrow x = {a^y}} \right] \\
  5\sqrt x - \sqrt {25x - 4} = 5 \\
  5\sqrt x - 5 = \sqrt {25x - 4} \]
Taking the terms common, we have
\[5\left( {\sqrt x - 1} \right) = \sqrt {25x - 4} \]
Squaring on both sides, we have
\[
  {5^2}{\left( {\sqrt x - 1} \right)^2} = {\left( {\sqrt {25x - 4} } \right)^2} \\
  25\left( {{{\left( {\sqrt x } \right)}^2} - 2\sqrt x + 1} \right) = 25x - 4 \\
  25\left( {x - 2\sqrt x + 1} \right) = 25x - 4 \\
  25x - 50\sqrt x + 25 = 25x - 4 \\
\]
Cancelling same terms on both sides, we have
\[
   - 50\sqrt x + 25 = - 4 \\
  50\sqrt x = 29 \\
  \sqrt x = \dfrac{{29}}{{50}} \\
\]
Squaring on both sides, we get
\[
  x = {\left( {\dfrac{{29}}{{50}}} \right)^2} \\
  \therefore x = {\left( {\dfrac{{29}}{{50}}} \right)^2} \\
\]
Thus, the value of \[x\] satisfying the equation \[{\log _{17}}{\log _5}\left( {5\sqrt x - \sqrt {25x - 4} } \right) = 0\] is \[{\left( {\dfrac{{29}}{{50}}} \right)^2}\]
And the value of \[50\sqrt x \] is 29.

Note: For log value to be zero, its upper part should be equal to zero, and for log value to be equal to 1, the upper part and base should be the same. Any integer raised to the power zero is equal to one. In this solution we have not taken zero as the value of \[x\], since it is not satisfying the given equation. In fact, if we put zero as the value of \[x\] it gives us infinity.