Question

pls solve it...step by step​

Answer 1

★Given:-

• $\sf log_2 x +log_4 x +log_{16}x = \frac{21}{4}$

★To find:-

• Value of x.

★Solution:-

Using the formula,

$\displaystyle \mapsto \underline{\boxed{\sf log_{a^n} N=\frac{1}{n}log_a N}}$

We can write,

$\displaystyle \implies \sf log_4 x = log_{2^2}x \\\\\implies \sf log_{16}x=log_{2^4}x$

Putting the the given equation,

$:\implies \sf log_2 x +log_{2^2}x +log_{2^4}x = \frac{21}{4}$

Using the base power rule of logarithm,the exponents can be separated from the bases of the logarithmic functions at base position.

$\displaystyle :\implies \sf log_2x+\frac{1}{2} log_2x + \frac{1}{4} log_2 x = \frac{21}{4}$

$\displaystyle :\implies \sf log_2 x \bigg(1+\frac{1}{2} +\frac{1}{4} \bigg) = \frac{21}{4}$

$:\implies \sf \dfrac{7}{4} log_2 x=\dfrac{21}{4}$

$\displaystyle :\implies \sf log_2x = \frac{21}{4} \times \frac{4}{7}$

$:\implies \sf log_2 x = 3$

$:\implies \sf x = 2^3$

$\displaystyle :\implies \underline{\boxed{\sf x = 8}}$

Hence, the value of x is 8.

Option(A) is correct.

_______________

Answer 2

Question:-

            1. If $\rm {log_2x+log_4x+log_16x=\frac{21}{4}}$, find x.

To find:-

• x

Solution:-

✔  $\rm log_b_c a=\frac{1}{c} log_b a$

✔ $\rm log_2 x + log_2^{2} x+log_2^{4} x = \frac{21}{4}$

✔ $\rm log_2 x+\frac{1}{2} log_2x+\frac{1}{4} log_2x=\frac{21}{4}$

✔ $\rm log_2 \times [1+\frac{1}{2} +\frac{1}{4}]=\frac{21}{4}$

✔ $\rm (log_2x)[\frac{4+2+1}{4} ]=\frac{21}{4}$

✔ $\rm (log_2x)(\frac{7}{4} )=\frac{21}{4}$

✔ $\rm log_2x=\frac{21}{4} \times\frac{4}{7} =\frac{21}{7} =3$

✔ $\rm log_2x=3$

✔ $\rm log_b a=c=a\neq b^{2}$

✔ $\rm log_2 x=3$

  ✔ x = $\bold\rm{2^{3}$

          = 8

Required answer:-

• 8