How do you solve $2 {log}_{2} (x) + {log}_{2} (1) = {log}_{2} (4)$ $2log_2 (x )+ log_2 (1)= log_2 (4)$ ?

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How do you solve $2 {log}_{2} (x) + {log}_{2} (1) = {log}_{2} (4)$ $2log_2 (x )+ log_2 (1)= log_2 (4)$ ?

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Answer 1 · 2015-12-24T07:32:35

Step by step explanation is given below.

Explanation:

$2 {log}_{2} (x) + {log}_{2} (1) = {log}_{2} (4)$ $2log_2(x)+log_2(1) = log_2(4)$

Let us use rules of logarithms

$1 . {log}_{b} (1) = 0$ $1. log_b (1) = 0$ $log (1)$ $log(1)$ to any base is zero
$2 . {log}_{b} (a^{n}) = n {log}_{b} (a)$ $2. log_b (a^n) = n log_b(a)$
$3 . {log}_{b} (A) = {log}_{b} (C) \Rightarrow A = C$ $3. log_b (A) = log_b (C) => A= C$

$2 {log}_{2} (x) + 0 = {log}_{2} (2^{2})$ $2log_2(x)+0 = log_2(2^2)$ by rule 1 and rewriting 4 as 2^2

$2 {log}_{2} (x) = 2 {log}_{2} (2)$ $2log_2(x) = 2log_2(2)$ by rule 2
${log}_{2} (x) = {log}_{2} (2)$ $log_2(x) = log_2(2)$ After dividing by 2 on both sides.
$x = 2$ $x = 2$ by rule 3.

The solution is $x = 2$ $x=2$

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How do you solve $2 {log}_{2} (x) + {log}_{2} (1) = {log}_{2} (4)$ $2log_2 (x )+ log_2 (1)= log_2 (4)$ ?

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Explanation:

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How do you solve $2 {log}_{2} (x) + {log}_{2} (1) = {log}_{2} (4)$ $2log_2 (x )+ log_2 (1)= log_2 (4)$ ?