# Thread: Need help solving another exponential problem.

1. ## Need help solving another exponential problem.

So I have the problem:

16^(d-4) = 3^(3-d)

(d-4)log(16) = (3-d)log(3)
(d-4)log16/log(3) = -(d-3)//Did not want 3-d so I made it d-3 with a negative in front.
-2.52(d-4) = d-3 //(-2.52 = log(16) / log(3)
-2.52d + 10.09 = d - 3
10.09 = -1.52d-3
13.09 = -1.52d
d = -8.594

But the answer in the back of the book says: d = 3.7162

Where did I go wrong?

2. You need to ADD $\displaystyle \displaystyle 2.52d$ to both sides...

3. Originally Posted by thyrgle
So I have the problem:

16^(d-4) = 3^(3-d)

(d-4)log(16) = (3-d)log(3)
(d-4)log16/log(3) = -(d-3)//Did not want 3-d so I made it d-3 with a negative in front.
-2.52(d-4) = d-3 //(-2.52 = log(16) / log(3)
-2.52d + 10.09 = d - 3
10.09 = -1.52d-3
13.09 = -1.52d
d = -8.594

But the answer in the back of the book says: d = 3.7162

Where did I go wrong? ... getting out your calculator too soon and messing with the signs.
$\displaystyle 16^{d-4} = 3^{3-d}$

$\displaystyle (d-4)\log{16} = (3-d)\log{3}$

$\displaystyle d\log{16} - 4\log{16} = 3\log{3} - d\log{3}$

$\displaystyle d\log{16} + d\log{3} = 4\log{16} + 3\log{3}$

$\displaystyle d(\log{16} + \log{3}) = 4\log{16} + 3\log{3}$

$\displaystyle d = \dfrac{4\log{16} + 3\log{3}}{\log{16} + \log{3}}$