# Test of Mathematics Solution Subjective 67 - Four Real Roots

This is a Test of Mathematics Solution Subjective 67 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

Describe the set of all real numbers x which satisfy 2 ${\log_{{2x+3}^x}}$ <1.

## Solution

2 ${\log_{{2x+3}^x}}$ < 1

Now x< 0 is not in the domain of logarithm.

So x> 0.

Now as x>0           2x+3 > 1.

So ${(2x+3)^a}$ > ${(2x+3)^b}$ for a>b
So 2 ${\log_{{2x+3}^x}}$ < 1 or ${(2x+3)^{\frac{1}{2}}}$ > x
or 2x+3 > ${x^2}$
or ${x^2}$ -2x -3 < 0
or -1 < x < 3

But x > 0 so set of all real number

x is 0 < x < 3.

# Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

## Problem - Trigonometry (SMO Test)

Find the value of $$(log_{\sqrt 2}(cos 20^\circ) + log_{\sqrt 2} (cos 40^\circ) + log_{\sqrt 2}(cos 80^\circ))^2$$

• 32
• 15
• 36
• 20

### Key Concepts

Trigonometry

Log Function

Challenges and Thrills - Pre - College Mathematics

## Try with Hints

This one is a very simple. We can start from here :

As all are in the function of log with $$\sqrt 2$$ as base so we can take it as common such that

$$log_{\sqrt 2}(cos 20 ^\circ . cos 40 ^\circ . cos 80^\circ)$$

Now as you can see we dont know the exact value of $$cos 20^\circ$$ or $$cos 40^\circ$$ or $$cos 80^\circ$$ values.

But theres a formula that we can use which is

cosA.cos B = $$\frac {1}{2} (cos (A+B) + cos (A-B))$$

Now try apply this formula in the above expression and try to solve.........

Now those who did not get the answer yet try this:

If we apply the formula in the expression mentioned in the last hint :

$$log_{\sqrt 2}(cos 20 ^\circ . \frac {1}{2}(cos 120 ^\circ . cos 40^\circ)$$

$$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{2} cos 40 ^\circ . cos 20^\circ)$$

$$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{4} (cos 60 ^\circ + cos 20^\circ)$$

We need to do the rest of the calculation.Try to do that .......................

Continue from the last hint:

$$log_{\sqrt 2} \frac {1}{8} = - \frac {log_{2} 8}{log_{2}(2^{\frac {1}{2}})} = -6$$

So squaring this answer = $$(-6)^2 = 36$$ ..........................(Answer)

# Geometric Progression- ISI Entrance B. Stat (Hons) 2003- Problem 3

## Geometric Progression

A geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. E.g., the height to which a ball rises in each successive bounce follows a geometric progression. The sequence 4, -2, 1,... is a Geometric Progression (GP) for which (-1/2) is the common ratio. We can use the concept to find an arbitrary term, a finite or infinite sum of the series, and apply them in various contexts, including some difficult problems.

## Try the problem

Suppose that the three distinct real numbers $$a,b \text{ and } c$$ are in G.P. and $$a+b+c=xb$$. Then

(A) $$-3<x<1 ;$$

(B) $$x>1$$ or $$x<-3 ;$$

(C) $$x>3$$ or $$x<-1 ;$$

(D) $$-1<x<3 ;$$

ISI entrance B. Stat. (Hons.) 2003 problem 3

Geometric Progression

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

If any three quantity are in GP then we have a relation between them, in this case $$a,b,c$$ are in G.P. so we have

$$b^2=ac$$ or $$b= \sqrt{ac}$$.

We also have $$a+b+c=xb$$ so

which is equal to $$a+c=b(x-1)$$
=$$\frac{a}{b}+\frac{c}{b}=(x-1)$$
since $$b=\sqrt{ac}$$ we will get
=$$\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}=(x-1)$$
=$$\sqrt{\frac{a}{c}}+\frac{1}{\sqrt{\frac{a}{c}}}=(x-1)$$
Let $$\sqrt{\frac{a}{c}}=k$$, then we have the form of $$k+\frac{1}{k}$$ which we know has a value either greater than 2 or less than -2.
so we can write
either, $$x-1 > 2$$ or $$x-1 < -2$$
and now you can easily get the answer.

So the answer is x>3 or x<-1.

# Logarithm ISI entrance B. Stat. (Hons.) 2003 problem 2

## Logarithm

The problem is based upon logarithm in which we find the value of an unknown quantity in an equation. And understanding whether the root or (value of x) is real or not and if yes, then how many real roots exist.

## Try the problem

The equation $$\log_3 x-\log_x 3 =2$$ has

(A) no real solution

(B) exactly one real solution

(C) exactly two real solution

(D) infinitely many real solution.

ISI entrance B. Stat. (Hons.) 2003 problem 2

Logarithm

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

we know when logarithm base and value are interchanged then the whole quantity is equal to the reciprocal of the previous logarithm.

i.e. $$\log_3 x$$ = $$\frac{1}{log_x 3}$$

Now we can assume the value of $$\log_3 x$$ is $$a$$ and the equation will reduce to $$a^2 -1/a=2$$ , or $$a^2-2a-1=0$$.

An now we can apply Sridharacharya's formula to find the valise(s) of x.

After solving we will get two values of a and they are

$$1\pm \sqrt{2}$$

And now these values will be equal to a or $$\log_3 x$$ , And from here we will get two values of x which are real.

So option (C) is the correct option.

# Dance of the Logarithm - AMC 10A 2014 Problem 25

In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication

AMC 10A 2014 Problem 25 (or 12A Problem 22) is a curious case of estimation gymnastics.

There are several treatments of this problem (search it online if you are into reading ‘solutions’). We will work through sequential hints as usual. Try the problem on your own instead of reading ‘solutions’.

- Sir Issac Newton

## Problem

The number $$5^{867}$$ is between $$2^{2013}$$ and $$2^{2014}$$. How many pairs of integers (m, n) are there such that $$1 \leq m \leq 2012$$ and $$5^n < 2^m < 2^{m+2} < 5^{n+1}$$ ?

• Logarithm

Also see

## Hint 1: $$log_2 5 \sim 2.33$$

This is actually very good estimate. Notice that $$2^2 < 5 < 2^{2 + \frac{1}{3}} = 2^{\frac{7}{3}}$$

How do we know this is true? It bears mark of a well established method of creating log tables since Napier’s time. Raise both sides to the power 3 and check if all is well. That is whether the following is true: $$(2^2)^3 < 5^3 < (2^{\frac{7}{3}})^3 = 2^7$$

This is not only true, $$5^3 =125$$ is very close to $$2^7 = 128$$

Hence $$log_2 5 \sim \frac{7}{3} \sim 2.33$$

## Hint 2: Logarithm is a monotonically increasing function

In other words, ‘taking log on all sides’ is a perfectly good way of preserving inequality. Hence take logarithm base 2 to have $$n \cdot \log_2 5 < m < m+2 < (n+1) \cdot \log_2 5$$

The left hand side inequality yields $$2.33n < m$$

The right hand piece yields $$m+2 < 2.33 (n+1 ) = 2.33n + 2.33 \Rightarrow m < 2.33n + 0.33$$

Combining these two we have $$2.33n < m < 2.33n + 0.33$$

## Hint 3: Integer in-between

2.33n and 2.33n + 0.33 are dangerously close! In order to squeeze a nice and round integer in between (that is m) we will need to be careful.

We really want the fractional part of 2.33n to be 0.67 or more. This works whenever n is a multiple of 3 from n= 1 to 33. (Check rigorously for n = 3k +2 and n= 3k+1, fractional part of 2.33n is not large enough. So adding 0.33 won't make 2.33n and 2.33n + 0.33 jump across an integer).

Each time we multiply by a greater number, the fractional part recedes by > 0.01. Hence by the time it is n = 30, it has receded enough to make room for the next power (3k+1 or 31)

The process repeats from next nine numbers of the form 3k+1 (31, 34, 37, 40, 43, 46, 49, 52, 55, 58) and then we slide over to 3k+2 and so on.

## Hint 4: Some of these 288 cases won't work.

Not all multiples of 3 are born equal. Ehm...

We want $$\log _2 5 \cdot n < m < \log_2 5 \cdot n + (\log_2 5 - 2)$$

That is $$\log_2 5^n < m < \log_2 \frac{5^{n+1}}{4}$$

Here is the observation: If I is an integer and I < x < x + t < I + 1 then 2x + t