Question

solve this ... in proper way .☺

Answer 1

Q : 24 (a) Solve for x :
$\frac{ log_{10}(x - 3) }{ log_{10}( {x}^{2} - 21 ) } = \frac{1}{2} \\ = > 2 log_{10}(x - 3) = log_{10}( {x}^{2} - 21 ) \\ = > log_{10}( {(x - 3)}^{2} ) = log_{10}( {x}^{2} - 21) \\ = > {(x - 3)}^{2} = {x}^{2} - 21 \\ = > 6x = 30 \\ = > x = 5$

Q : 24 (b)
$log( log(x) ) + log( log( {x}^{3} ) - 2 ) = 0$
where base of log is 10 everywhere .
And. 24 (b)
$log( log(x) \times (log( {x}^{3} ) - 2)) = 0 \\ = > log(x) \times (3 log(x) - 2) = {10}^{0} \\ = > 3 {( log(x) )}^{2} - 2 log(x) - 1 = 0 \\ = > ( 3log(x) + 1)( log(x) - 1) = 0 \\ = > log(x) = 1 \: \: \: or \: log(x) = \frac{ - 1}{3}$
We rejected log(x) = -1/3 as then log(log (x)) will not be defined as input becomes negative.
Therefore,
$log(x) = 1 \\ = > x = {10}^{1} = 10$

Q : 24 (c)
$log_{x}(2) \times log_{2x}(2) = log_{4x}(2) \\ = > \frac{ log(2) }{ log(x) } \times \frac{ log(2) }{ log(2x) } = \frac{ log(2) }{ log(4x) } \\ = > log(2) \times log(4x) = log(x) \times log(2x) \\ = > log(2) \times (2 log(2) + log(x) ) \\ \: \: \: \: \: = log(x) \times ( log(2) + log(x) ) \\ = > 2 {( log(2) )}^{2} = {( log(x)) }^{2} \\ = > log(x) = + \sqrt{2} log(2) \: \: or \: \\ \: \: \: \: \: \: - \sqrt{2} log(2) \\ = > log(x) = log( {2}^{ \sqrt{2} } ) \: \: or \: \\ log( {2}^{ - \sqrt{2} } ) \\ = > x = {2}^{ \sqrt{2} } \: \: or \: {2}^{ - \sqrt{2} }$

Q : 24 (d)
${5}^{ log(x) } + 5 {x}^{ log(5) } = 3 \\ where \: base \: of \: log \: is \: a > 0$
And.
${5}^{ log_{a}(x) } + 5 \times {5}^{ log_{a}(x) } = 3 \\ = > 6 \times {5}^{ log_{a}(x) } = 3 \\ = > {5}^{ log_{a}(x) } = \frac{1}{2} \\ = > x = {2}^{ \frac{ - log(a) }{ log(5) } } = {2}^{ - log_{5}(a) }$