How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2 x+log_4 x=log_2 5$ ?

Question

How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2 x+log_4 x=log_2 5$ ?

1 Answer

Impact of this question

1962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-02T13:09:22

$x = 5^{\frac{2}{3}}$ $x = 5^(2/3)$

Explanation:

Note that

${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log _2 x+log_4x = log_2 5$
$\Rightarrow \frac{ln x}{ln 2} + \frac{ln x}{{ln 2}^{2}} = \frac{ln 5}{ln 2}$ $implies (lnx)/(ln2) + (lnx)/(ln2^2) = (ln5)/(ln2)$
$\Rightarrow \frac{ln x}{ln 2} + \frac{ln x}{2 ln 2} = \frac{ln 5}{ln 2}$ $implies (lnx)/(ln2) + (lnx)/(2ln2) = (ln5)/(ln2)$
$\Rightarrow \frac{ln x}{ln 2} (1 + \frac{1}{2}) = \frac{ln 5}{ln 2}$ $implies(lnx)/(ln2)(1+1/2) = (ln5)/(ln2)$
$\Rightarrow ln x = \frac{2}{3} ln 5$ $implies lnx = 2/3 ln5$
$\Rightarrow ln x = {ln 5}^{\frac{2}{3}}$ $implies lnx = ln5^(2/3)$
$\Rightarrow x = 5^{\frac{2}{3}}$ $implies x = 5^(2/3)$

Science

Math

Humanities

... and beyond

How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2 x+log_4 x=log_2 5$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve log_2 x+log_4 x=log_2 5log2x+log4x=log25log_2 x+log_4 x=log_2 5?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve ${log}_{2} x + {log}_{4} x = {log}_{2} 5$ $log_2 x+log_4 x=log_2 5$ ?