Question

Give answer if you know only :


$\displaystyle \text{How many solutions the equation log_x216+log_{2x}64=3 has?}$


Option are :

a . One irrational solution

b . No solution is a prime number

c . Two real solution

d . One integral solution


No spam need contain quality answer with great explanation.

Answer 1

Answer:

$log_x{2} {}^{16} +log_{2x}64=3$

It can be written as

$log_{x}(2 {}^{2 {}^{4} } ) + log_{2x}(2 {}^{6} ) = 3 \\ \\ \frac{4}{2} log_{x}(2) + 6 log_{2x}(2) = 3 \\ \\ 2 \times log_{x}(2) + 6 \times log_{2x}(2) = 3$

As We Know log_a(b)=1/log_b(a)

$2 \times \frac{1}{log_{2}(x)} + 6 \times \frac{1}{ log_{2}(2 \times x) } = 3 \\ \\ \frac{2}{ log_{2}(x) } + \frac{6}{ log_{2}(2) + log_{2}(x) } \\ \\ \frac{2}{ log_{2}(x) } + \frac{6}{1 + log_{2}(x) }$

Let log_2(x)=t

$\frac{2}{t } + \frac{6}{1 + t} \\ \\ \frac{2(1 + t) + 6t}{t(t + 1)} = 3 \\ \\ 2 + 2t + 6t = 3t(t + 1) \\ \\ 2 + 8t = 3 {t}^{2} + 3t \\ \\ {3t}^{2} + 3t - 8t - 2 \\ \\ {3t}^{2} - 5t - 2 \\ \\ {3t}^{2} - 6t + t - 2 \\ \\ 3t(t - 2) + 1(t - 2) \\ \\ (3t + 1)(t - 2)$

Now

3t+1=0

3t=-1

t=-1/3

T=log_2(x)

$log_{2}(x) = \frac{1}{3} \\ \\ x = 2 {}^{ \frac{1}{3} } \\ \\ x = \sqrt[3]{2}$

Now

t-2=0

t=2

$log_{2}(x) = 2 \\ \\ x = {2}^{2} \\ \\ x = 4$

We get Two Value of X=³√2,4

Now They are Real Number

So Option C is Correct

C.Two Real solution

Answer 2

Answer:

C.two real solutions

Thanks .