How do you solve $2log_6 4-1/4log_6 16=log_6 x$ ?

Question

3958 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T10:08:23

$x = 8$

We have: $2 log_(6)(4) - frac(1)(4) log_(6)(16) = log_(6)(x)$

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

$Rightarrow 2 log_(6)(4) - frac(1)(4) log_(6)(4^(2)) = log_(6)(x)$

$Rightarrow 2 log_(6)(4) - frac(2)(4) log_(6)(4) = log_(6)(x)$

$Rightarrow 2 log_(6)(4) - frac(1)(2) log_(6)(4) = log_(6)(x)$

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

$Rightarrow (2 - frac(1)(2))log_(6)(4) = log_(6)(x)$

$Rightarrow frac(3)(2) log_(6)(4) = log_(6)(x)$

$Rightarrow log_(6)(4^(frac(3)(2))) = log_(6)(x)$

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

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

$Rightarrow 6^(log_(6)(4^(frac(3)(2)))) = 6^(log_(6)(x))$

$Rightarrow 4^(frac(3)(2)) = x$

$therefore x = 8$

How do you solve 2log_6 4-1/4log_6 16=log_6 x2log_6 4-1/4log_6 16=log_6 x?