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# Logarithms and some other stuff

I have a few - a lot, actually - of doubts with logarithms and some other stuff. Anyone up? I'd be thankful even if you could answer one of these.

Q1. $\log _{ 5 }{ 120 } +\left( x-3 \right) -2\cdot \log _{ 5 }{ \left( 1-{ 5 }^{ x-3 } \right) } =-\log _{ 5 }{ \left( 0.2-{ 5 }^{ x-4 } \right) }$

Q2. Find the sum of all the solutions of the equation $\large{{ 3 }^{ { \left( \log _{ 9 }{ x } \right) }^{ 2 }-{ \frac { 9 }{ 2 } \log _{ 9 }{ x } }+5 }=3\sqrt { 3 } }$

Q3. Let $$a$$, $$b$$, $$c$$ and $$d$$, be positive integers such that $$\log_{a}{b}=\frac{3}{2}$$ and $$\log_{c}{d}=\frac{5}{4}$$. If $$(a-c)=9$$, find the value of $$(b-d)$$.

Q4. If $\log_{3x}{45}=\log_{4x}{40\sqrt{3}}$ then find the characteristic of $$x^3$$ to the base 7.

Q5. Find $$x$$ satisfying the equation $\log^{2}{\left(1+\frac{4}{x}\right)} + \log^{2}{\left(1-\frac{4}{x+4}\right)}=2\log^{2}{\left(\frac{2}{x-1}-1\right)}$

Q6. Find the real solutions to the system of equations $\log_{10}{(2000xy)}-\log_{10}{x}\cdot\log_{10}{y}=4$ $\log_{10}{(2yz)}-\log_{10}{y}\cdot\log_{10}{z}=1$ $\log_{10}{(zx)}-\log_{10}{z}\cdot\log_{10}{x}=0$

Q7. Solve : $\log _{ 3 }{ \left( \sqrt { x } +\left| \sqrt { x } -1 \right| \right) =\log _{ 9 }{ \left( 4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| \right) } }$

Q8. Prove that $\large{{ 2 }^{ \left( \sqrt { \log _{ a }{ \sqrt [ 4 ]{ ab } +\log _{ b }{ \sqrt [ 4 ]{ ab } } } } -\sqrt { \log _{ a }{ \sqrt [ 4 ]{ \frac { b }{ a } } } +\log _{ b }{ \sqrt [ 4 ]{ \frac { a }{ b } } } } \right) \cdot \sqrt { \log _{ a }{ b } } }=\begin{cases} 2 \\ { 2 }^{ \log _{ a }{ b } } \end{cases}\begin{matrix} { \text{ if }\quad b\geq a>1 } \\ \text{ if }\quad 1<b<a \end{matrix}}$

Q9. Find the value of $\frac { 1 }{ \sin { 3\alpha } } \left[ \sin ^{ 3 }{ \alpha } +\sin ^{ 3 }{ \left( \frac { 2\pi }{ 3 } +{ \alpha } \right) +\sin ^{ 3 }{ \left( \frac { 4\pi }{ 3 } +{ \alpha } \right) } } \right]$

Q10. Find the value of $$a$$ for which the equation $\lvert x^2-4x+3\rvert = x+a$ has exactly three distinct real roots.

Q11. Find the number of terms of the longest geometric progression that can be obtained from the set $$(100,~101,\dots ,~1000)$$. (I think the question does not consider $$r=1$$, because the answer given for this one is $$6$$.)

Q12. If $$p(x)=ax^2+bx+c$$ and $$q(x)=-ax^2+dx+c$$, where $$ac\neq0$$, then prove that $$p(x)q(x)$$ has at least two real roots. (I think it should be $$q(x)=-ax^2+bx+c$$, but this is what the sheet says.)

Q13. If $$x$$ and $$y$$ are real numbers such that $x^2+2xy-y^2=6$ find the minimum value of $$\left(x^2+y^2\right)^2$$.

Q14. If the product $(\sin1^{\circ})(\sin3^{\circ})(\sin5^{\circ})(\sin7^{\circ})\dots(\sin89^{\circ})=\frac{1}{2^n}$ then find the value of $$[n]$$. (where $$[y]$$ denotes greatest integer less than or equal to $$y$$.)

Note by Omkar Kulkarni
2 years, 1 month ago

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For question no. 1 , I got answer as x=1.

- 8 months, 2 weeks ago

- 2 years, 1 month ago

Thank you!

- 2 years, 1 month ago

For 10, I get 2 solutions, $$a = -1, a = -0.75$$.

- 2 years, 1 month ago

Could you explain how?

- 2 years, 1 month ago

Uh, I did it graphically.

Draw $$x^2 - 4x + 3$$ ( Parabola cutting x -axis at (1,0), (3,0)). Invert the part under the x -axis to get the graph of $$|x^2 - 4x + 3|$$. Now look at the family of lines which have slope of 45 deg and see which of them cut the function at 3 points.

There should be two lines. One line cuts the graph at (1,0) and therefore must be $$y = x - 1$$.

The other is tangent to the graph between x =1 and x =3. The slope of the line is $$1$$. The slope of the function between $$x = 1$$ and $$x =3$$ is $$-2x +4$$. Since the slopes are equal, $$x = 3/2$$. Putting in the function $$y = 3/4$$. Line which satisfies this is $$x = y - 0.75$$.

- 2 years, 1 month ago

Oh right. Thanks!

- 2 years, 1 month ago

For Q11.

Should the answer be $$\infty$$? Let common ratio=1

- 2 years, 1 month ago

I believe the question doesn't consider that, because the answer given here is $$6$$.

- 2 years, 1 month ago

Please mention that. Otherwise other people may get confused like me :P

- 2 years, 1 month ago

Yeah, good idea :P

- 2 years, 1 month ago

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