How do you write these expressions as single logarithms? 1. $4 log 2 + log 6$ $4 log 2 + log 6$ 2. $3 {log}_{2} 6 - 2$ $3log_2 6-2$ 3. $5 log 3 - 2 log 8$ $5 log 3 - 2 log 8$

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Answer 1 · 2018-02-17T21:49:54

$4 log 2 + log 6$ $4 log 2 + log 6$
to start off, write the expression using exponents.
${log 2}^{4} + log 6$ $log 2^4 + log 6$
since we are adding, we multiply
$log (2^{4} \cdot 6)$ $log(2^4*6)$
$= log (16 \cdot 6)$ $=log(16*6)$
$= log 96$ $=log 96$
$3 {log}_{2} 6 - 2$ $3 log_2 6 - 2$
for this one, we need to make sure the bases are the same so that we are able to combine them.
${log}_{2} 6^{3} - {log}_{2} 2^{2}$ $log_2 6^3 - log_2 2^2$
since are subtracting, we divide
${log}_{2} (\frac{6^{3}}{2^{2}})$ $log_2(6^3/2^2)$
$= {log}_{2} (\frac{216}{4})$ $=log_2(216/4)$
$= {log}_{2} 54$ $=log_2 54$
$5 log 3 - 2 log 8$ $5 log 3 - 2 log 8$
${log 3}^{5} - {log 8}^{2}$ $log 3^5 - log 8^2$
$log (\frac{3^{5}}{8^{2}})$ $log(3^5/8^2)$
$= log (\frac{243}{64})$ $=log(243/64)$

Answer 2 · 2018-02-17T22:01:49

When adding two logs together you multiply
$log (2) + log (6)$ $log(2)+log(6)$
$log (2 \cdot 6)$ $log(2*6)$
If there is a number behind the log it becomes an exponent
$4 log (2) + log (6)$ $4log(2)+log(6)$
$log (2^{4} \cdot 6)$ $log(2^4*6)$
$log (96)$ $log(96)$
Refer to answer below
When you subtract logs you divide
$5 log (3) - 2 log (8)$ $5log(3)−2log(8)$
$log (\frac{3^{5}}{2^{8}})$ $log(3^5/2^8)$
$log (\frac{243}{64})$ $log(243/64)$

How do you write these expressions as single logarithms? 1. 4log2+log64 log 2 + log 6 2. 3log26−23log_2 6-2 3. 5log3−2log85 log 3 - 2 log 8