Solving an Exponential Equation using Logs: Base-2 HW help (1 question)

[Doctor Beaker]

Solve for T.
The Problem: 5*2^-3t=45

 

[Steven G]

2^-3t =9
t=9/2^-3
t =9*8
t=72

wow t=72 on my 7,200th post

 

[Doctor Beaker]

2^-3t =9
t=9/2^-3
t =9*8
t=72

wow t=72 on my 7,200th post

I thought we would have to convert 2^-3t=9 into logarithmic form.

 

[Deleted member 4993]

Solve for T.
The Problem: 5*2^-3t=45

Is your problem statement:

\(\displaystyle 5 * 2^{-3} * t = 45\)

or

\(\displaystyle 5 * 2^{-3 * t} = 45\)

 

[Doctor Beaker]

Is your problem statement:

\(\displaystyle 5 * 2^{-3} * t = 45\)

or

\(\displaystyle 5 * 2^{-3 * t} = 45\)

The bottom one.

 

[Otis]

I thought we would have to convert 2^(-3t)=9 into logarithmic form.

Hi doc. I think Jomo might be playing a stunt, instead of correcting your notation.

… The Problem: 5*2^(-3t)=45

I've inserted the required grouping symbols, above.

Start by taking the base-2 logarithm of each side, and then apply the following property of logarithms (on the left-hand side).

log(n^c) = c · log(n)

At the end, you may use the property for changing bases (to convert your answer from base-2 to another base, if you desire).

Does that help?

?

 

[Doctor Beaker]

I'm confused. I got to this part so far: t=log2(9)/-3

 

[Otis]

I'm confused. I got to this part so far: t=log2(9)/-3

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 exact answer in terms of the natural logarithm, then you may apply the change-of-base formula.

log_b(x) = ln(x)/ln(b)

?

 

[Doctor Beaker]

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.

?

But where does the -1/3 come from? I can't remember how to get that.

 

[pka]

The bottom one is \(5\cdot 2^{-3t}=45\) or \(2^{-3t}=9\)
Using plain logarithms: \(-3t\log(2)=2\log(3)\)
So \(t=\dfrac{-2}{3}\dfrac{\log(3)}{\log(2)}\)

 

[Otis]

But where does the -1/3 come from? I can't remember how to get that.

Dividing by -3 is the same as multiplying by -1/3.

PS: An alternative notation for posting log_b(x) is log[b](x)

?

 

[Doctor Beaker]

Dividing by -3 is the same as multiplying by -1/3.

PS: An alternative notation for posting log_b(x) is log(x)

?

Thank you for your help.

 

[Otis]

Thank you for your help.

My pleasure. Did you see what pka did? He applied the same property to log₂(9) as we did earlier to log₂(2^-3t).

log₂(9) may be viewed as log₂(3²).

That's why we can write t = -2/3 · log₂(3)

?

 

[pka]

PS: An alternative notation for posting log_b(x) is log[b](x)

Perhaps you don't know that in most current mathematics instruction one common logarithm is used.
If anyone who might need say \(\log_2(3)\) that is equal to \(\dfrac{\log(3)}{\log(2)}\).
Those of us who following Leonard Gillman think the idea of logarithm should be simplified.