Rewriting an exponential function in terms of e

[Adi]

In the solutions to my practice exam there was a step where 1.2^t was rewritten as e^ln(1.2)xt. I don't understand how this works so there is clearly a serious gap in my knowledge. Is someone able to explain this to me step by step? Thanks

 

[Dr.Peterson]

In the solutions to my practice exam there was a step where 1.2^t was rewritten as e^ln(1.2)xt. I don't understand how this works so there is clearly a serious gap in my knowledge. Is someone able to explain this to me step by step? Thanks

This is a standard process that you should become familiar with.

It relies on the fact that the exponential function is the inverse of the log, so that [MATH]b^{log_b\left(x\right)} = x[/MATH].

In this case, we can replace 1.2 with [MATH]e^{ln(1.2)}[/MATH], so we have [MATH]1.2^t = \left(e^{ln(1.2)}\right)^t = e^{ln(1.2)\cdot t}[/MATH].

 

[Adi]

Thanks a lot Dr Peterson. I've been putting different numbers into the calculator and rewriting them to familiarise myself with the process

 

[JeffM]

Let's see why it is true.

[MATH]0 < y = b^{log_b(x)} \implies log_b(y) = log_b(b^{log_b(x)}) \implies[/MATH]
[MATH]log_b(y) = log_b(x) * log_b(b) = log_b(x) * 1 \implies[/MATH]
[MATH]log_b(y) = log_b(x) \implies y = x \implies[/MATH]
[MATH]x > 0 \implies x = b^{log_b(x)}.[/MATH]

 

[Adi]

Haha thanks JeffM. You're a legend!

 

[HallsofIvy]

Dr. Peterson wrote: "In this case, we can replace 1.2 with \(\displaystyle e^{ln(1.2)^t}\), so we have \(\displaystyle 1.2^t=(e^ln(1.2))^t=e^{ln(1.2)⋅t}\)."

I would have written this slightly differently. \(\displaystyle ln(a^b)= bln(a)\) so
\(\displaystyle 1.2^t= e^{ln(1.2^t)}= e^{t ln(1.2)}\).

 

[Otis]

… e^ln(1.2)xt …

? \(\;\) Here's a tip: Don't use symbol x as a multiplication sign. Instead, type a dot, asterisk or use grouping symbols. (Beginning with algebra, symbol x represents a number.)

e^ln(1.2)∙t

e^ln(1.2)*t

e^ln(1.2)(t)

?

 

[Dr.Peterson]

Dr. Peterson wrote: "In this case, we can replace 1.2 with \(\displaystyle e^{ln(1.2)^t}\), so we have \(\displaystyle 1.2^t=(e^ln(1.2))^t=e^{ln(1.2)⋅t}\)."

I would have written this slightly differently. \(\displaystyle ln(a^b)= bln(a)\) so
\(\displaystyle 1.2^t= e^{ln(1.2^t)}= e^{t ln(1.2)}\).

What I said got a little mangled there; I said this:

We can replace 1.2 with [MATH]e^{ln(1.2)}[/MATH], so we have [MATH]1.2^t = \left(e^{ln(1.2)}\right)^t = e^{ln(1.2)\cdot t}[/MATH].

The two approaches actually aren't so much "slightly different ways to write the same thing", but are different approaches that sort of work from the inside out vs. from the outside in. Both are perfectly valid, and somewhat a matter of taste (or of what fits best with your ways of thinking).

Mine is a replacement on the inside, replacing the base 1.2 with an exponential form, so that raising this to a power multiplies the exponents.

Halls' way applies the same fact, that [MATH]u = e^{ln(u)}[/MATH], to the entire expression [MATH]1.2^t[/MATH] rather than just to 1.2, and simplifying inward.

If one of these makes more sense to you, stick with that one!