## Practice questions on Indices Surds and Logarithms

### 1. Practice questions on Indices Surds and Logarithms

Question  01:  Which of the following is the largest?
(1) $${6^{{2^{48}}}}$$
(2) $${25^{{2^{47}}}}$$
(3) $${256^{{2^{46}}}}$$
(4) $${6561^{{2^{45}}}}$$
Choice (1)

We need to compare
$${6^2}^{48},\,\,\,{25^2}^{47} = {5^{{2^1} \times {2^{47}}}} = {5^2}^{48},\,$$
$$\,{256^2}^{46} = {4^{4 \times {2^{46}}}} = {4^{{2^{48}}}} = {2^{{2^{49}}}},$$
$${6561^{{2^{45}}}} = {3^{8 \times {2^{45}}}} = {3^{{2^3} \times {2^{45}}}} = {3^{{2^{48}}}}$$
$$\,{65536^{{2^{44}}}} = {2^{16 \times {2^{44}}}} = {2^{{2^4} \times {2^{44}}}} = {2^{{2^{48}}}}$$

Question  02:  If $${a^x} = {b^y} = {c^z}$$ and $$abc = 1$$, then what is the value of $$xy + yz + zx$$
(1) 0
(2) $$x + y + z$$
(3) 1
(4) $${x^2} + {y^2} + {z^2}$$
Choice (1)

Let $${a^x} = {b^y} = {c^z} = k$$, then
$$a = {k^{1/x}},\;\,b = {k^{1/y}},\,c = {k^{1/z}}$$
Given that $$abc = 1$$, hence $${k^{1/x}}{k^{1/y}}{k^{1/z}} = 1 = {k^0}$$
$$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$ or $$xy + yz + zx = 0$$

Question  03:  Find the relation in $$a$$, $$b$$ and $$c$$, given that $${\left( {0.25} \right)^a} = {\left( {20} \right)^b} = {\left( 5 \right)^c}$$
(1) $$\frac{2}{c} - \frac{2}{b} = \frac{1}{a}$$
(2) $$\frac{2}{c} + \frac{2}{b} = \frac{1}{a}$$
(3) $$\frac{1}{c} + \frac{1}{b} = \frac{1}{a}$$
(4) $$\frac{1}{c} - \frac{1}{b} = \frac{1}{a}$$
Choice (4)

Given that $${\left( {0.25} \right)^a} = {\left( {20} \right)^b} = {\left( 5 \right)^c} = k$$(say)
$$\Rightarrow \frac{1}{4} = {k^{1/a}},\,20 = {k^{1/b}},\,5 = {k^{1/c}}$$
Also $$\left( {\frac{5}{{20}}} \right) = \frac{1}{4} \Rightarrow \frac{{{k^{1/c}}}}{{{k^{1/b}}}} = {k^{1/a}}$$
$$\Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{a}$$

Question  04:  Simplify $$\left( {1 + 17} \right)\left( {1 + {{17}^2}} \right)\left( {1 + {{17}^4}} \right)\left( {1 + {{17}^8}} \right)...\left( {1 + {{17}^{{2^{100}}}}} \right)$$
(1) $$\frac{{{{17}^{{2^{200}}}} - 1}}{{16}}$$
(2) $$\frac{{{{17}^{{2^{101}}}} - 1}}{{16}}$$
(3) $$\frac{{{{17}^{{2^{102}}}} - 1}}{{16}}$$
(4) None of these
Choice (2)

Multiplying and dividing by (17 – 1), then product can be simplified to:
$$\frac{{(17 - 1)(17 + 1)({{17}^2} + 1)....({{17}^{{2^{100}}}} + 1)}}{{(17 - 1)}}$$
$$\frac{{({{17}^{{2^{101}}}} - 1)}}{{(17 - 1)}} = \frac{{({{17}^{{2^{101}}}} - 1)}}{{16}}$$

Question  05:  If $${(4.7)^x} = {(0.047)^y} = 1000$$, then $$\frac{1}{x} - \frac{1}{y}$$ is:
(1) 44930
(2) 44929
(3) 44928
(4) 44960
Choice (4)

Given that, $${4.7^x} = 1000$$, hence
$$4.7 = {1000^{\frac{1}{x}}}$$
Similarly, $$0.047 = {1000^{\frac{1}{y}}}$$
Dividing the first equation by the second equation, we have
$$\frac{{4.7}}{{0.047}} = {1000^{\frac{1}{x} - \frac{1}{y}}}$$
$$\Rightarrow 100 = {1000^{\frac{1}{x} - \frac{1}{y}}}$$
$$\Rightarrow {10^2} = {10^{3\left( {\frac{1}{x} - \frac{1}{y}} \right)}}$$
Hence $$\frac{1}{x} - \frac{1}{y} = \frac{2}{3}$$

Question  06:  The number of distinct integral values of a satisfying the equation $${2^{2a}} - {3.2^{a + 2}} + {2^5} = 0$$ is:
(1) 0
(2) 1
(3) 2
(4) 3
Choice (3)

Assuming $${2^a} = x$$, the given equation becomes
$${x^2}-12x + 32 = 0$$ $$\Rightarrow x = 4,8$$
Thus $$a = 2,\,\,3$$. There are 2 solutions.

Question  07:  Find the value of $$x$$ if $${x^{{x^3}}} = 3$$.
(1) 3
(2) $$\sqrt 3$$
(3) 0.33
(4) $${3^{1/3}}$$
Choice (4)

Let $${x^3} = a$$$$\Rightarrow x = {a^{1/3}}$$ …(1)
From the given relation $${x^a} = 3 \Rightarrow x = {3^{1/a}}$$
From the equation (1), $${a^{1/3}} = {3^{1/a}}$$
$$\Rightarrow {a^a} = {3^3}$$ or $$a = 3$$and $$x = {3^{\frac{1}{3}}}$$

Question  08:  If $${3^{324}} = {m^m}$$, where $$m$$ is a natural number, then find the value of $$\sqrt m$$
(1) $$\sqrt 3$$
(2) 3
(3) 9
(4) 81
Choice (3)

$${3^{324}} = {\left( {{3^4}} \right)^{81}}$$$$= {81^{81}}$$
Hence $$m = 81$$ or $$\sqrt m = 9$$

Question  09:  For all possible integers $$n$$ satisfying $$2.25 \le 2 + {2^{n + 2}} \le 202$$, the number of integer values of $$n$$ is
(1) 2
(2) 8
(3) 9
(4) 10
Choice (4)

The given inequality can be written as
$$0.25 \le {2^{n + 2}} \le 200$$
We know that $$0.25 = \frac{1}{4} = {2^{ - 2}}$$ and $${2^8} = 256$$
$$\Rightarrow - 2 \le n + 2 \le 7\;{\rm{or}}\; - 4 \le n \le 5$$
Hence n can take 10 values.

Question  10:  Find the biggest number among $${2^{\frac{1}{2}}},\,{3^{\frac{1}{3}}},\,{5^{\frac{1}{4}}}$$ and $${6^{\frac{1}{5}}}$$.
(1) $${5^{\frac{1}{4}}}$$
(2) $${3^{\frac{1}{3}}}$$
(3) $${6^{\frac{1}{5}}}$$
(4) $${2^{\frac{1}{2}}}$$
Choice (1)

Let us compare the first three numbers and take 12 powers of each so that the power of each becomes a natural number.
$${\left( {{2^{\frac{1}{2}}}} \right)^{12}},\,{\left( {{3^{\frac{1}{3}}}} \right)^{12}},\,{\left( {{5^{\frac{1}{4}}}} \right)^{12}}$$= $${2^6},\,{3^4},\,{5^3}$$
We see that $${5^3}$$is the biggest among the three numbers.
Now we can compare $${5^{\frac{1}{4}}}\;{\rm{and}}\;{6^{\frac{1}{5}}}$$ using the same approach.

Question  11:  If $$x =\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$ and $$y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}$$, then find the value of $${x^3} + {y^3}$$
(1) 122
(2) 244
(3) 488
(4) None of these
Choice (3)

Given that x = $$\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$= $$\frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{5 - 3}}$$
= $$4 + \sqrt {15}$$, similarly, y = $$4 - \sqrt {15}$$
Now $$xy = 1$$ and $${x^2} + {y^2} = 62$$
$$\Rightarrow {x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} + {y^2}--xy} \right)$$
$$\Rightarrow {x^3} + {y^3} = \left( 8 \right)\left( {62--1} \right) = 488.$$

Question  12:  Find the value of $$\frac{{\sqrt {26 - 15\sqrt 3 } }}{{\left( {5\sqrt 2 - \sqrt {38 + 5\sqrt 3 } } \right)}}$$
(1) 1
(2) $$\frac{1}{{\sqrt 3 }}$$
(3) $$\sqrt 3$$
(4) 3
Choice (2)

Multiplying numerator and denominator by $$\sqrt 2$$
$$\Rightarrow \sqrt {52 - 2\sqrt {25 \times 27} } \Rightarrow \frac{{\sqrt {27} - \sqrt {25} }}{{10 - (\sqrt {75} + \sqrt 1 )}} = \frac{{3\sqrt 3 - 5}}{{9 - 5\sqrt 3 }}$$
Rationalizing, we get $$\frac{1}{{\sqrt 3 }}$$.

Question  13:  Find the simplified value of $$\left( {\frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }}} \right)$$
(1) $$\sqrt 2$$
(2) $$\sqrt 3$$
(3) 2
(4) $$\sqrt 6$$
Choice (2)

The given surd can be written as
$$\left( {\frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }}} \right) = \frac{{3 + \sqrt 6 }}{{\sqrt {3 + \sqrt 2 } }} = \sqrt 3$$.

Question  14:  If $$n$$ is a positive integer such that $$\left( {\sqrt[5]{{10}}} \right){\left( {\sqrt[5]{{10}}} \right)^2}{\left( {\sqrt[5]{{10}}} \right)^3}....{\left( {\sqrt[5]{{10}}} \right)^n} > 999$$, then the smallest value of $$n$$ is
(1) 3
(2) 4
(3) 5
(4) 6
Choice (3)

The number is $${10^{\frac{1}{5}}}{10^{\frac{2}{5}}}{......10^{\frac{n}{5}}} > 999$$
$$\Rightarrow {10^{\frac{1}{5}\left( {1 + 2 + 3 + ... + n} \right)}} > 999$$
$$\Rightarrow {10^{\frac{{n(n + 1)}}{{10}}}} > 999$$
Since 1000 > 999, hence $$\frac{{n(n + 1)}}{{10}} \ge 3$$
Or $$n(n + 1) \ge 30$$ $$\Rightarrow n \ge 5$$

Question  15:  What is the value of $$\frac{1}{3}{\log _{10}}125 - 2{\log _{10}}4 + {\log _{10}}32 - 2$$
(1) 1
(2) 0
(3) – 1
(4) $$\frac{1}{2}$$
Choice (3)

The simplified value is:
$${\log _{10}}{125^{1/3}} - {\log _{10}}{4^2} + {\log _{10}}32 - {\log _{10}}100$$
$$= {\log _{10}}\left( {\frac{{5 \times 32}}{{16 \times 100}}} \right) = {\log _{10}}\left( {\frac{1}{{10}}} \right) = - 1$$

Question  16:  If $$\sqrt[m]{{\sqrt[n]{{\sqrt[n]{m}}}}} = \,\,\,n$$ then find $${\log _n}m$$
(1) mn2
(2) m – 2
(3) $$\frac{2}{m}$$
(4) $$\frac{1}{{n - 2}}$$
Choice (1)

$$\sqrt[m]{{\sqrt[n]{{\sqrt[n]{m}}}}} = \,n$$$$\Rightarrow \,{m^{\frac{1}{{m{n^2}}}}} = n \Rightarrow m = {n^{mn}}^2$$
$$\Rightarrow {\log _n}m = {\log _{_{^n}}}{n^{m{n^2}}} = m{n^{^{_2}}}$$

Question  17:  If $${\log _3}({\log _3}({\log _3}x)) = 1,$$then the value of $$x$$ is:
(1) $${27^9}$$
(2) 27
(3) $${3^{18}}$$
(4) $${9^{18}}$$
Choice (1)

From the given relation,
$${\log _3}({\log _3}x) = {3^1} = 3$$
$$\Rightarrow {\log _3}x = {3^3} = 27$$
Hence $$x = {3^{{3^3}}} = {3^{27}} = {({3^3})^9} = {27^9}$$

Question  18:  Find $${\log _{30}}8$$if $${\log _{30}}3 = \alpha$$and $${\log _{30}}5 = \beta$$
(1) $$3(1 – \alpha – \beta)$$
(2) $$3(1 – \alpha + \beta)$$
(3) $$3( 1+ \alpha + \beta)$$
(4) $$2(1 – \alpha – \beta)$$
Choice (1)

$${\log _{30}}8 = 3{\log _{30}}2 = 3{\log _{30}}\left( {\frac{{30}}{{15}}} \right)$$
= $$3(1 - {\log _{30}}15) = 3(1 - {\log _{30}}3 - {\log _{30}}5)$$
= $$3(1--\alpha --\beta )$$

Question  19:  The simplified value of $${\log _{49}}343 - {\log _{343}}49$$
(1) $$\frac{6}{5}$$
(2) $$\frac{5}{6}$$
(3) 1
(4) None of these
Choice (2)

We know that $${\log _{{a^m}}}x = \frac{1}{m}{\log _a}x$$
Hence $${\log _{49}}343 = {\log _{{7^2}}}343 = \frac{1}{2}{\log _7}343 = \frac{3}{2}$$
Also $${\log _{343}}49 = \frac{1}{{{{\log }_{49}}343}} = \frac{2}{3}$$
Required answer = $$\frac{3}{2} - \frac{2}{3} = \frac{5}{6}$$

Question  20:  Which of the following is a correct order, where $$a$$is more than 1.
(1) $${\log _2}a > {\log _3}a > {\log _e}a > {\log _{20}}a$$
(2) $${\log _3}a > {\log _e}a > {\log _4}a > {\log _{10}}a$$
(3) $${\log _2}a > {\log _e}a > {\log _4}a > {\log _{16}}a$$
(4) $${\log _e}a > {\log _\pi }a > {\log _3}a > {\log _4}a$$
Choice (3)

Value of the logarithm decreases when base increases for any value of a (> 1).

Question  21:  Find the value of $$100x$$ if $$5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \log \frac{1}{{\sqrt {1 - {x^2}} }}$$
(1) 10
(2) 99
(3) $$\frac{{99}}{{100}}$$
(4) 100
Choice (2)

We can write $$5 = {\log _{10}}{10^5}$$, hence the equation can be written as
$$\log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)$$
$$\Rightarrow \log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {(1 + x)(1 - x)} }}} \right)$$
$$\Rightarrow {10^5}{\left( {\sqrt {1 - x} } \right)^5} = 1$$ or $$\sqrt {1 - x} = \frac{1}{{10}}$$
$$\Rightarrow x = \frac{{99}}{{100}}\;{\rm{or}}\;100x = 99$$

Question  22:  If $${\rm{lo}}{{\rm{g}}_{\rm{2}}}\left[ {{\rm{3\; + \;lo}}{{\rm{g}}_{\rm{3}}}\{ {\rm{4\; + \;lo}}{{\rm{g}}_{\rm{4}}}\left( {x - 1} \right)\} } \right] - {\rm{2\; = \;0}}$$ then $$4x$$ equals
(1) 2
(2) 3
(3) $$\frac{5}{4}$$
(4) 5
Choice (4)

Given that $${\log _2}[3 + {\log _3}\{ 4 + {\log _4}(x - 1)\} ] = 2$$
$$\Rightarrow 3 + {\log _3}\{ 4 + {\log _4}(x - 1)\} = {2^2} = 4$$
$$\Rightarrow {\log _3}\{ 4 + {\log _4}(x - 1)\} = 1$$
$$\Rightarrow 4 + {\log _4}(x - 1) = 3$$
$$\Rightarrow {\log _4}(x - 1) = - 1$$
Hence $$x - 1 = \frac{1}{4}$$ or $$4x = 5$$

Question  23:  If $${\log _4}5 = \left( {{{\log }_4}y} \right)\left( {{{\log }_6}\sqrt 5 } \right)$$, then $$y$$ equals
(1) 25
(2) 36
(3) $$5\sqrt 5$$
(4) 75
Choice (2)

The equation can be written as
$$\frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\log \sqrt 5 }}{{\log 6}}} \right)$$
$$\Rightarrow \frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\frac{1}{2}\log 5}}{{\log 6}}} \right)$$
$$\Rightarrow \frac{1}{2}\log y = \log 6\;{\rm{or}}\;y = 36$$

Question  24:  If $${\log _2}3 = m$$and $${\log _3}5 = n$$, then the value of $${\log _{30}}45$$in terms of m and n
(1) $$\frac{{m + mn}}{{1 + m + mn}}$$
(2) $$\frac{{2m + mn}}{{1 + n + mn}}$$
(3) $$\frac{{2m + mn}}{{1 + m + mn}}$$
(4) None of these
Choice (3)

$${\log _{30}}45 = \frac{{{\rm{lo}}{{\rm{g}}_{\rm{2}}}45}}{{{{\log }_2}30}} = \frac{{2{{\log }_2}3 + {{\log }_2}5}}{{1 + {{\log }_2}3 + {{\log }_2}5}}$$ …(1)
Since $${\log _2}5 = \frac{{{{\log }_3}5}}{{{{\log }_3}2}} = \frac{n}{{1/m}} = mn$$
Putting these values in (1), we get the answer.

Question  25:  Simplified value of $${\log _{ab}}x$$
(1) $$\frac{{{{\log }_a}x + {{\log }_b}x}}{{{{\log }_a}x{{\log }_b}x}}$$
(2) $$\frac{{{{\log }_a}x + {{\log }_b}x}}{{{{\log }_a}x - {{\log }_b}x}}$$
(3) $$\frac{{{{\log }_a}x{{\log }_b}x}}{{{{\log }_a}x + {{\log }_b}x}}$$
(4) None of these
Choice (3)

$${\log _{ab}}x = \frac{1}{{{{\log }_x}ab}} = \frac{1}{{{{\log }_x}a + {{\log }_x}b}}$$
$$= \frac{1}{{\frac{1}{{{{\log }_a}x}} + \frac{1}{{{{\log }_b}x}}}}$$= $$\frac{{{{\log }_a}x{{\log }_b}x}}{{{{\log }_a}x + {{\log }_b}x}}$$.

Question  26:  If $$\frac{{\log \alpha }}{{\beta - \gamma }} = \frac{{\log \beta }}{{\gamma - \alpha }} = \frac{{\log \gamma }}{{\alpha - \beta }}$$, then which of the following is true?
(1) $${\alpha ^2}{\beta ^2}{\gamma ^2} = 2$$
(2) $${\alpha ^\beta }{\beta ^\gamma }{\gamma ^\alpha } = 1$$
(3) $${\alpha ^\alpha }{\beta ^\beta }{\gamma ^\gamma } = 1$$
(4) None of these
Choice (3)

Let $$\frac{{\log \alpha }}{{\beta - \gamma }} = \frac{{\log \beta }}{{\gamma - \alpha }} = \frac{{\log \gamma }}{{\alpha - \beta }} = k$$, then
$${\log _x}\alpha = k(\beta - \gamma ) \Rightarrow \alpha = {x^{k(\beta - \gamma )}}$$
Similarly, $$\beta = {x^{k(\gamma - \alpha )}}$$, $$\gamma = {x^{k(\alpha - \beta )}}$$
Now $${\alpha ^\alpha }{\beta ^\beta }{\gamma ^\gamma } = {x^{k\alpha (\beta - \gamma ) + k\beta (\gamma - \alpha ) + k\gamma (\alpha - \beta )}} = {x^0} = 1$$.

Question  27:  If $$\frac{{x(y + z - x)}}{{\log x}} = \frac{{y(z + x - y)}}{{\log y}} = \frac{{z(x + y - z)}}{{\log z}}$$, then
(1) $${x^x}{y^y}{z^z} = 1$$
(2) $${x^x} = {y^y} = {z^z}$$
(3) $${x^y}{y^x} = {y^z}{z^y} = {x^z}{z^x}$$
(4) $${x^y}{y^z}{z^x} = 1$$
Choice (3)

Suppose $$\frac{{x(y + z - x)}}{{\log x}} = \frac{{y(z + x - y)}}{{\log y}} = \frac{{z(x + y - z)}}{{\log z}} = a$$
$$\Rightarrow \log x = \frac{x}{a}(y + z - x),\log y = \frac{y}{a}(x + z - y)$$
and $$\log z = \frac{z}{a}(x + y - z)$$
Now $$y\log x + x\log y = \frac{{2xyz}}{a}$$
$$z\log y + y\log z = \frac{{2xyz}}{a}$$and $$z\log x + x\log z = \frac{{2xyz}}{a}$$
or $${x^y}{y^x} = {y^z}{z^y} = {x^z}{z^x}$$

Question  28:  Find the value of y if
$${\log _2}x + {\log _4}y + {\log _4}z = 4$$
$${\log _3}y + {\log _9}z + {\log _9}x = 4$$
$${\log _5}z + {\log _{25}}x + {\log _{25}}y = 4$$
(1) 2.7
(2) 9
(3) 7.29
(4) None of these
Choice (3)

The first equation can be written as
$${\log _{{2^2}}}{x^2} + {\log _4}y + {\log _4}z = 4$$
$$\Rightarrow {\log _4}({x^2}yz) = 4\;{\rm{or}}\;{x^2}yz = {4^4}$$ ….. (1)
Similarly, second the third equations can be written as,
$${y^2}xz = {9^4}$$ ….. (2)
and $${z^2}yx = {25^4}$$ ….. (3)
Multiplying all the three equations, we have
$${x^4}{y^4}{z^4} = {(4 \times 9 \times 25)^4} \Rightarrow xyz = 900$$ ….. (4)
Now using equation (2) and (4),
$$y = \frac{{{9^4}}}{{900}} = 7.29$$

Question  29:  For a real number$$a$$, if $$\frac{{{{\log }_{15}}a + {{\log }_{32}}a}}{{{{\log }_{15}}a{{\log }_{32}}a}} = 4$$ then $$a$$ must lie in the range
(1) $$4 < a < 5$$
(2) $$3 < a < 4$$
(3) $$a > 5$$
(4) $$2 < a < 3$$
Choice (1)

The equation can be written as
$$\frac{1}{{{{\log }_{32}}a}} + \frac{1}{{{{\log }_{15}}a}} = 4$$
$$\Rightarrow {\log _a}32 + {\log _a}15 = 4$$
$$\Rightarrow {\log _a}(15 \times 32) = 4$$$$\Rightarrow {a^4} = 15 \times 32 = 480$$
Thus $$4 < a < 5$$

Question  30:  If $$y$$ is a negative number such that $${2^{{y^2}{{\log }_3}5}} = {5^{{{\log }_2}3}}$$, then $$y$$ equals
(1) $${\log _2}\left( {\frac{1}{3}} \right)$$
(2) $${\log _2}\left( {\frac{1}{5}} \right)$$
(3) $$- {\log _2}\left( {\frac{1}{3}} \right)$$
(4) $$- {\log _2}\left( {\frac{1}{5}} \right)$$
Choice (1)

Taking log both sides
$${y^2}{\log _3}5\log 2 = {\log _2}3\log 5$$
$$\Rightarrow \frac{{{y^2}\log 5}}{{\log 3}} \times \log 2 = \frac{{\log 3}}{{\log 2}} \times \log 5$$
$$\Rightarrow {y^2} = {\left( {\frac{{\log 3}}{{\log 2}}} \right)^3}$$
$$\Rightarrow y = - \frac{{\log 3}}{{\log 2}} = - {\log _2}3 = {\log _2}\left( {\frac{1}{3}} \right)$$