How do you solve $16^{d - 4} = 3^{3 - d}$ $16^(d-4)=3^(3-d)$ ?

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Answer 1 · 2015-10-23T16:41:47

$d = \frac{log 1769472}{log 48}$ $d=(log1769472)/(log48)$

Using laws of exponents we may write this equation as

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

$therefore(16xx3)^d=3^3*16^4$

Now taking the log on both sides and simplifying we get

$dlog48=log1769472$

$therefore d=(log1769472)/(log48)$

How do you solve 16^(d-4)=3^(3-d)16d−4=33−d16^(d-4)=3^(3-d)?