# How do you solve 16^(d-4)=3^(3-d)?

Oct 23, 2015

$d = \frac{\log 1769472}{\log 48}$

#### Explanation:

Using laws of exponents we may write this equation as

${16}^{d} \cdot {16}^{- 4} = {3}^{3} \cdot {3}^{- d}$

$\therefore {\left(16 \times 3\right)}^{d} = {3}^{3} \cdot {16}^{4}$

Now taking the log on both sides and simplifying we get

$\mathrm{dl} o g 48 = \log 1769472$

$\therefore d = \frac{\log 1769472}{\log 48}$