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

Oct 19, 2016

$d = 3.7162$

#### Explanation:

As ${16}^{d - 4} = {3}^{3 - d}$, taking log to the base $10$ on both sides, we get

$\left(d - 4\right) \log 16 = \left(3 - d\right) \log 3$

or $d \times \log 16 - 4 \log 16 = 3 \times \log 3 - d \times \log 3$

or $d \left(\log 16 + \log 3\right) = 3 \times \log 3 + 4 \log 16$

or $d = \frac{3 \times \log 3 + 4 \log 16}{\log} 48$

= $\frac{3 \times 0.4771 + 4 \times 1.2041}{1.6812}$

= $\frac{1.4313 + 4.8164}{1.6812}$

= $\frac{6.2477}{1.6812} = 3.7162$