Doctor Beaker
New member
- Joined
- Jun 22, 2020
- Messages
- 6
Solve for T.
The Problem: 5*2^-3t=45
The Problem: 5*2^-3t=45
I thought we would have to convert 2^-3t=9 into logarithmic form.2^-3t =9
t=9/2^-3
t =9*8
t=72
wow t=72 on my 7,200th post
Is your problem statement:Solve for T.
The Problem: 5*2^-3t=45
The bottom one.Is your problem statement:
\(\displaystyle 5 * 2^{-3} * t = 45\)
or
\(\displaystyle 5 * 2^{-3 * t} = 45\)
Hi doc. I think Jomo might be playing a stunt, instead of correcting your notation.I thought we would have to convert 2^(-3t)=9 into logarithmic form.
I've inserted the required grouping symbols, above.… The Problem: 5*2^(-3t)=45
That's a correct answer, doc.I'm confused. I got to this part so far: t=log2(9)/-3
But where does the -1/3 come from? I can't remember how to get that.That's a correct answer, doc.
I would write it like this:
t = -1/3 · log2(9)
If you would like to evaluate a decimal approximation, or if you would like to express the answer in terms of the natural logarithm (or base-10), then you may apply the change-of-base formula.
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Dividing by -3 is the same as multiplying by -1/3.But where does the -1/3 come from? I can't remember how to get that.
Thank you for your help.Dividing by -3 is the same as multiplying by -1/3.
PS: An alternative notation for posting logb(x) is log(x)
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My pleasure. Did you see what pka did? He applied the same property to log2(9) as we did earlier to log2(2-3t).Thank you for your help.
Perhaps you don't know that in most current mathematics instruction one common logarithm is used.PS: An alternative notation for posting logb(x) is log[b](x)