12
Q:

# The least value of expression $\inline 2\log_{10}x-\log_{x}(1/100)$ for x>1 is:

 A) 2 B) 3 C) 4 D) 5

Explanation:

$\inline&space;2\log_{10}x-\log_{x}\frac{1}{100}=2\log_{10}x-\frac{\log_{10}10^{-2}}{\log_{10}x}$

$\inline&space;=2\log_{10}x+\frac{2}{\log_{10}x}$

$\inline&space;=2\left&space;[&space;\log_{10}x-\frac{1}{\log_{10}x}&space;\right&space;]$

Since       x >1     $\inline&space;\Rightarrow&space;\log_{10}x>0$

But since $\inline&space;AM\geq&space;GM$

$\therefore$  $\inline&space;\frac{\log_{10}x+\frac{1}{\log_{10}x}}{2}\geq&space;\sqrt{\log_{10}x\times&space;\frac{1}{\log_{10}x}}$

$\inline&space;\Rightarrow$   $\inline&space;\log_{10}x+\frac{1}{\log_{10}x}\geq&space;2$

$\Rightarrow$   $\inline&space;2\left&space;[&space;\log_{10}x+\frac{1}{\log_{10}x}&space;\right&space;]\geq&space;4$

For x= 10,  $\inline&space;2\left&space;[&space;\log_{10}x+\frac{1}{\log_{10}x}&space;\right&space;]\geq&space;4$

Hence the least value of $\inline&space;\left&space;[&space;\log_{10}x-\log_{x}\frac{1}{100}&space;\right&space;]$ is 4

Q:

If $\inline \fn_jvn \log _{7}2$ = m, then $\inline \fn_jvn \log _{49}28$ is equal to ?

 A) 1/(1+2m) B) (1+2m)/2 C) 2m/(2m+1) D) (2m+1)/2m

Explanation:

$\inline \fn_jvn \log _{49}28 = \frac{1}{2\log _{7}(7x4)}$

= $\inline \fn_jvn \frac{1}{2}(1+\log _{7}4)$
= $\inline \fn_jvn \frac{1}{2}+\frac{1}{2}(2\log _{7}2)$
$\inline \fn_jvn \frac{1}{2}+\log _{7}2$
$\inline \fn_jvn \frac{1}{2}+m$

$\inline \fn_jvn \frac{(1+2m)}{2}$.

3 151
Q:

If $\fn_jvn \small a^{2}+b^{2}=c^{2}$ , then $\inline \fn_jvn \frac{1}{\log_{c+a}b} + \frac{1}{\log_{c-a}b}= ?$

 A) 1 B) 2 C) 4 D) 8

Explanation:

Given $\fn_jvn \small a^{2}+b^{2}=c^{2}$

Now $\inline \fn_jvn \frac{1}{\log _{c+a}b}+\frac{1}{\log _{c-a}b}$ = $\inline \fn_jvn \log _{b}(c+a)+\log _{b}(c+a)$

$\inline \fn_jvn \log _{b}(c^{2}-a^{2})$

$\inline \fn_jvn \log _{b}b^{2}=2\log _{b}b = 1$

4 178
Q:

If log 64 = 1.8061, then the value of log 16 will be (approx)?

 A) 1.9048 B) 1.2040 C) 0.9840 D) 1.4521

Explanation:

Given that, $\inline \fn_jvn \small \log 64=1.8061$

$\inline \fn_jvn \small i.e. \log 4^{3}=1.8061$

$\inline \fn_jvn \small \Rightarrow 3\log 4=1.8061$

$\inline \fn_jvn \small \Rightarrow \log 4=0.6020$

$\inline \fn_jvn \small \Rightarrow 2\log 4=1.2040$

$\inline \fn_jvn \small \Rightarrow \log 4^{2}=1.2040$

$\inline \fn_jvn \small \Rightarrow \log 16=1.2040(approx)$

7 502
Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

 A) 100 m B) 125 m C) 150 m D) 175 m

Explanation:

$\inline \fn_jvn Time = \frac{sum\: of\: length\: of\: the\: two\: train}{Difference\: in\: speeds}$

$\inline \fn_jvn 20 = \frac{(100+x)}{25/2}$

$\inline \fn_jvn \Rightarrow x= 150\: m$

8 783
Q:

For $\inline \fn_jvn x\epsilon N, x>1$$\inline \fn_jvn P = \log_{x}(X+1)$ and  $\inline \fn_jvn Q = \log_{(X+1)}(X+2)$, then which one of the following is correct?

 A) P < Q B) P = Q C) P > Q D) can't be determined

Explanation:

$\inline \fn_jvn \frac{k}{l}>\frac{k+1}{l+1}$ for (k,l) > 0 and  k > l

Let     k = x+1    and   l = x

$\inline \fn_jvn \therefore \frac{x+1}{x}>\frac{(x+1)+1}{(x)+1}$

$\inline \fn_jvn \because (x+1)>x$

$\inline \fn_jvn \therefore \frac{log(x+1)}{logx}>\frac{log(x+2)}{log(x+1)}$

$\inline \fn_jvn \Rightarrow \log_{x}(x+1)>\log_{x+1}(x+2)$