# University Maths Solution

Maths Question
Question 1

Given that

$p=\frac{1}{{{\log }_{a}}(2-\sqrt{3})}+\frac{1}{{{\log }_{b}}\left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right)}$

and ${{(2+\sqrt{3})}^{p}}=\frac{1}{12}$

where a, b > 0, find ab

Question 2

$\text{Show that }{{\log }_{a}}{{(a+b)}^{2}}=2+{{\log }_{a}}\left( 1+\frac{2b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}} \right)$

Question 3

${{\log }_{4}}(x+2){{\log }_{x}}2=1$

Question 4

${{\log }_{x}}4+{{\log }_{4}}x=2\tfrac{1}{2}$

Question 5

$\text{Show that }\frac{\log \sqrt{27}+\log \sqrt{8}-\log \sqrt{125}}{\log 5-\log 6}=-\frac{3}{2}$

Question 6

\begin{align} & \text{Show without using tables that } \\ & (i)\text{ }2{{\log }_{10}}6+3{{\log }_{10}}2={{\log }_{10}}288 \\ & (ii)\text{ }{{\log }_{10}}19=\tfrac{1}{2}(2{{\log }_{10}}6+1) \\ & \text{ Hence, show that }{{\log }_{10}}17=\tfrac{1}{2}(2{{\log }_{10}}6+3{{\log }_{10}}2) \\ & (iii)\text{ Deduce from these results that }\sqrt{5}\approx \frac{38}{17} \\\end{align}

Question 7

$6({{9}^{\tfrac{1}{x}}})-13({{6}^{\tfrac{1}{x}}})+5({{4}^{\tfrac{1}{x}}})=0$

Question 8

\begin{align} & \text{If }a={{x}^{2n}},\text{ }b={{y}^{-n}},\text{ }c={{x}^{\tfrac{n}{2}}}{{y}^{\tfrac{2}{3}}} \\ & \text{Find} \\ & (i)\text{ }abc\text{ } \\ & (ii)\text{ }\tfrac{ab}{c}\text{ } \\ & (iii)\text{ }{{\log }_{e}}a+{{\log }_{e}}bc\text{ } \\ & (iv)\text{ lo}{{\text{g}}_{e}}abc\text{ } \\ & (v)\text{ }{{\log }_{e}}\tfrac{b}{c}-{{\log }_{e}}c\text{ } \\ & (vi)\text{ }\frac{{{\log }_{e}}a-{{\log }_{e}}b}{{{\log }_{e}}c} \\ \end{align}

Question 9

\begin{align} & \text{If }x={{\log }_{a}}\left( \frac{a}{a+1} \right),\text{ }y={{\log }_{a}}\left( \frac{a}{a-1} \right),\text{ }z={{\log }_{a}}\left( \frac{{{a}^{2}}}{{{a}^{2}}-1} \right) \\ & \text{Evaluate} \\ & (i)\text{ }x+y-z\text{ }(ii)\text{ }x+2y-z\text{ (iii)}\,\text{ }z-x-y\text{ }(iv)\text{ }2x+y-z \\\end{align}

Question 10

\begin{align} & \text{If }x{{\log }_{16}}N={{\log }_{2}}N,\text{ Find the value of }x.\text{ } \\ & \text{Hence find }N\text{ if }{{\log }_{5}}N+{{\log }_{25}}N=6 \\ \end{align}

Question 11

$\text{Without using table, show that }\frac{\log {{8}^{\tfrac{2}{3}}}+\log {{729}^{\tfrac{1}{3}}}-\log {{125}^{\tfrac{2}{3}}}}{\log 6-\log 5}=2$

Question 12

$\text{If }{{x}^{2}}+{{y}^{2}}=2xy,\text{ prove that }\log (x+y)=\log 2+\tfrac{1}{2}\log x+\tfrac{1}{2}\log y$

Question 13

\begin{align} & \text{Let }x={{(a+b)}^{\tfrac{1}{4}}}+{{(a-b)}^{\tfrac{1}{4}}},\text{ }y={{(a+b)}^{\tfrac{1}{4}}}-{{(a-b)}^{\tfrac{1}{4}}}\text{ and } \\ & {{a}^{2}}-{{b}^{2}}={{p}^{4}};\text{ prove }{{x}^{2}}-4p-{{y}^{2}}=0 \\\end{align}

Question 14

\begin{align} & \text{By putting},\alpha =\log a,\text{ }\beta =\log b,\text{ }\gamma =\log c\text{ in the identity } \\ & \text{show that }\alpha (\beta -\gamma )+\beta (\gamma -\alpha )+\gamma (\alpha -\beta )=0 \\\end{align}

Question 15

Find the relation between a and b not involving logarithms if ${{\log }_{9}}a=2+{{\log }_{3}}b$

Question 16

Given that $1+{{\log }_{3}}p={{\log }_{27}}q$ obtain a relation between p and q without involving logarithms

Question 17

Given that ${{\log }_{8}}(p+2)+{{\log }_{8}}q=r-\tfrac{1}{3}$ and ${{\log }_{2}}(p-2)-lo{{g}_{2}}q=2r+1$ , show that ${{p}^{2}}=4+{{32}^{r}}$.

Question 18

If ${{\log }_{3}}(x-6)=2y$ and ${{\log }_{2}}(x-7)=3y$ show that ${{x}^{2}}-13x+42={{72}^{y}}$ Given that y = 1. Find the possible value(s) of x

Question 19

Simplify $\frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}$

Question 20

Solve the equation ${{\log }_{2}}({{x}^{2}}-5x-10)=2$

Question 21

If ${{p}^{2}}=qr$ show that ${{\log }_{q}}p+{{\log }_{r}}p=2{{\log }_{q}}p{{\log }_{r}}p$

Question 22

If ${{a}^{2}}+{{b}^{2}}=23ab$ show that $\log a+\log b=2\log \left( \frac{a+b}{5} \right)$

Question 23

Solve the equation ${{\log }_{10}}({{x}^{2}}+9)-2{{\log }_{10}}x=1$

Question 24

Find x if  ${{\log }_{x}}8-{{\log }_{{{x}^{2}}}}16=1$

Question 25

Find x if ${{\log }_{x}}3+{{\log }_{3}}x=2.5$

Question 26

If $2{{\log }_{y}}x+2{{\log }_{x}}y=5$ show that ${{\log }_{y}}x$ is either ½ or 2. Hence find the all pair of values of x and y which satisfy above and the equation xy =27