Question

I am giving uhhh challenge.
$\rm \implies log_{4}(x) + log_{2}(x) = 6$
Solve for x. ​

Answer 1

Step-by-step explanation:

i sent it as2 pics .........

Answer 2

Required Answer:-

Given:

• $\rm log_{4}(x) + log_{2}(x) = 6$

To find:

• The value of x.

Solution:

We have,

$\rm \implies log_{4}(x) + log_{2}(x) = 6$

We know that,

$\rm \leadsto log_{x}(y) = \dfrac{ log(y) }{ log(x) }$

So,

$\rm \implies \dfrac{ log(x) }{ log(4) } + \dfrac{ log(x) }{ log(2) } = 6$

$\rm \implies \dfrac{ log(x) }{ log(2^{2}) } + \dfrac{ log(x) }{ log(2) } = 6$

We know that,

$\rm \leadsto log( {x}^{y} ) = y log(x)$

So,

$\rm \implies \dfrac{ log(x) }{2 log(2) } + \dfrac{ log(x) }{ log(2) } = 6$

$\rm \implies \dfrac{ log(x) + 2 log(x) }{2 log(2) } = 6$

$\rm \implies \dfrac{3 \log(x) }{2 \log(2) } = 6$

$\rm \implies \dfrac{ \log(x) }{ \log(2) } = 6 \times \dfrac{2}{3}$

$\rm \implies \dfrac{ \log(x) }{ \log(2) } = 4$

$\rm \implies log_{2}(x) = 4$

$\rm \implies x = {2}^{4}$

$\rm \implies x = 16$

Hence, the values of x is 16.

Answer:

• The value of x is 16.