How do you solve $2 {log}_{6} 4 - \frac{1}{4} {log}_{6} 16 = {log}_{6} x$ $2log_6 4-1/4log_6 16=log_6 x$ ?

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Answer 1 · 2018-03-15T10:08:23

$x = 8$ $x = 8$

We have: $2 {log}_{6} (4) - \frac{1}{4} {log}_{6} (16) = {log}_{6} (x)$ $2 log_(6)(4) - frac(1)(4) log_(6)(16) = log_(6)(x)$

Let's begin this by expressing $- \frac{1}{4} {log}_{6} (16)$ $- frac(1)(4) log_(6)(16)$ in terms of an argument of $4$ $4$ :

$\Rightarrow 2 {log}_{6} (4) - \frac{1}{4} {log}_{6} (4^{2}) = {log}_{6} (x)$ $Rightarrow 2 log_(6)(4) - frac(1)(4) log_(6)(4^(2)) = log_(6)(x)$

$\Rightarrow 2 {log}_{6} (4) - \frac{2}{4} {log}_{6} (4) = {log}_{6} (x)$ $Rightarrow 2 log_(6)(4) - frac(2)(4) log_(6)(4) = log_(6)(x)$

$\Rightarrow 2 {log}_{6} (4) - \frac{1}{2} {log}_{6} (4) = {log}_{6} (x)$ $Rightarrow 2 log_(6)(4) - frac(1)(2) log_(6)(4) = log_(6)(x)$

Then, we can subtract the like terms on the left-hand side of the equation:

$\Rightarrow (2 - \frac{1}{2}) {log}_{6} (4) = {log}_{6} (x)$ $Rightarrow (2 - frac(1)(2))log_(6)(4) = log_(6)(x)$

$\Rightarrow \frac{3}{2} {log}_{6} (4) = {log}_{6} (x)$ $Rightarrow frac(3)(2) log_(6)(4) = log_(6)(x)$

$\Rightarrow {log}_{6} (4^{\frac{3}{2}}) = {log}_{6} (x)$ $Rightarrow log_(6)(4^(frac(3)(2))) = log_(6)(x)$

Now, both sides of the equation are in terms of the logarithm of base $6$ $6$ .

We can cancel these logarithms out by exponentiating both sides by $6$ $6$ :

$\Rightarrow 6^{{log}_{6} (4^{\frac{3}{2}})} = 6^{{log}_{6} (x)}$ $Rightarrow 6^(log_(6)(4^(frac(3)(2)))) = 6^(log_(6)(x))$

$\Rightarrow 4^{\frac{3}{2}} = x$ $Rightarrow 4^(frac(3)(2)) = x$

$therefore x = 8$

How do you solve 2log_6 4-1/4log_6 16=log_6 x2log64−14log616=log6x2log_6 4-1/4log_6 16=log_6 x?