SOLUTION: How do I solve with x as power?
2^x+4 = 4(16^x)
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Question 339875: How do I solve with x as power?
2^x+4 = 4(16^x)
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How do I solve with x as power?
2^(x+4) = 4(16^x)
---
2^(x+4) = 4(2^4x)
2^(x+4) = 2^2(2^4x)
2^(x+4) = 2^(4x+2)
x+4 = 4x+2
3x = 2
x = 2/3
==============
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Use logarithms. Any base will do in the final analysis, but the arithmetic will be neater in this case if you use base 2.
First, I will make the assumption that you mean:
}\ =\ 4\left(16^x\right))
as opposed to:
)
First note that 
Hence your equation becomes:
}\ =\ 4\left(2^{4x}\right))
Now take the base 2 logarithm of both sides:
}\right)\ =\ \log_2\left(4\left(2^{4x}\right)\right))
Use the following laws of logarithms to simplify the equation:
\ =\ n\log_b(x))
 + \log_b(y) = \log_b(xy))
\log_2\left(2\right)\ =\ \log_2\left(4\right)\ +\ 4x\,\cdot\,\log_2\left(2\right))
Next note that \ =\ 1\ \forall\ b\ \in\ \mathbb{R},\ b\ >\ 0)
So:
\ +\ 4x)
But also recall that:
 \ \ \Rightarrow\ \ b^y = x)
So we can say:
\ =\ 2)
and then:
Which you can solve by ordinary means.
John
My calculator said it, I believe it, that settles it
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