SOLUTION: log2 18+log2 x-log2 3
the twos are lowercase i am having a hard time doing these problems.
log2 x+log2(x+3)-log2x^2
solve x: -5=log2x
e^5x+1=40 solve
7log+log2 32
i
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: log2 18+log2 x-log2 3
the twos are lowercase i am having a hard time doing these problems.
log2 x+log2(x+3)-log2x^2
solve x: -5=log2x
e^5x+1=40 solve
7log+log2 32
i
Log On
Question 47490: log2 18+log2 x-log2 3
the twos are lowercase i am having a hard time doing these problems.
log2 x+log2(x+3)-log2x^2
solve x: -5=log2x
e^5x+1=40 solve
7log+log2 32
if f(x)=log2x find f(16)
I have over a hundred problems like these for an exam if i can get someone to show me how to do this by monday will be great
thank you Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
log2 x+log2(x+3)-log2x^2
Using the log laws you can rewrite this expression as:
log[x*(x+3)/x^2
=log2 [(x^2+3x)/x^2]
---------------------------
solve x: -5=log2x
2^(-5)=x
x=1/32
----------------------------
e^5x+1=40 solve
e^(5x+1)=40
Take the natural log of both sides to get:
5x+1 = ln40
5x+1 = 3.68887945...
5x=2.68887945...
x=0.53777589...
----------------------------
7log+log2 32
This problem is garbled
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if f(x)=log2x find f(16)
f(16)=log2 16 = 4
-----------------------
Cheers,
Stan H.
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