solving log equations with no decimal approximations

[jhawk555]

I am struggling at grasping this concept and wanted to check my work to solve the folllowing equation.

1. 9^5x-3=78

This is what I did:

ln9^5x-3 = ln78

(5x-3)ln9=ln78

added 3 to both sides and divided 5 and ln9 thus getting:

x= ln78 + 3 over 5ln9 = 4.144495694 or 4.1445



2. 2e^0.25x=8

I started by dividing 2 on both sides getting:

e^0.25x = 4

lne^0.25x = ln4

0.25x=4

I then divided both sides by 0.25 getting ln4/0.25

any and all help is greatly appreciated. [/tex]

 

[jwpaine]

9^(5x-3)=78

this is not a natural base exponential, so you would not use ln.

log is the inverse of an exponential, so bring the exponent down and log both sides.

(5x-3)log(9) = log(78)

divide both sides by log(9) add 3 to both sides, divide both sides by 5 to get x by itselt.

5x/5-3+3 = ((log(78)/log(9))+3)/5

x = approx .996565



You are correct on the second one!

2e^0.25x=8
divide both sides by 2

e^0.25x = 4

ln both sides

0.25x = ln(4)
x = ln(4)/0.25
x = approx 5.54518

 

[jhawk555]

Thank you. I am getting these things mixed up.

 

[pka]

9^(5x-3)=78 this is not a natural base exponential, so you would not use ln.

I totally disagree with that idea!
I like most working mathematicians think that the natural logarithm is the only one that is to be ever used in mathematics. I know that physicist and chemist disagree with that. But then they just use mathematics. I know from bitter experience the difficulty trying to teach complex analysis to physicist and electrical engineering students. In such a course there is but one logarithm, log(x). It is based on the Euler constant, e. Do those students ever get confused!

 

[jwpaine]

pka, you have said this before, but I still don't fully understand.

9^(5x-3)=78

why would you use anything but base 10 for this? To use loge^() would require a base change, sense this is not really a natural exponential.

Can you please explain how you would solve this with your "one and only log base e"

I don't understand.


Thanks,
John.

 

[stapel]

To use loge^() would require a base change, sense this is not really a natural exponential.

How is log-base-e not the natural logarithm? And why would a base-change be required before using a calculator, since most (all?) scientific and graphing calculators have an "LN" key?

Please reply with clarification. Thank you.

Eliz.

 

[pka]

9^(5x-3)=78, why would you use anything but base 10 for this?

What is the point of the question? What is the point of any question involving an equation? Isn’t it the idea in this case to find a value of x the ‘satisfies’ the equation?
That is, 9 raised to the power, five times that number minus three, equals 78.
\(\displaystyle \L 9^{5x - 3} = 78\quad \Rightarrow \quad 5x - 3 = \frac{{\ln (78)}}{{\ln (9)}}\quad \Rightarrow \quad x = \frac{{\ln (78)}}{{5\ln (9)}} + \frac{3}{5}.\).
There we have found such a number. I have also posted a screen shot of this same problem done with a well known computer algebra system.

Please note the form of the answer MathCad gave. Also see that the answer I gave is numerically equivalent.

Post Script to Ms. Stapel’s.
As a mater of fact the logarithm base e is universally known as the natural logarithm.

 

[jwpaine]

When I said "this is not really a natural exponential" I was revering to the problem.

I can see if you had e^x = 5 you would ln both sides so x = ln(5)

but if you had 5^x = 10, pke is saying that he disagrees with my statement about only using ln for solving natural base exponents.

I don't understand how you would solve 5^x = 10 using ln. I thought ln (because it is the inverse of the natural base exponent) would only be used when solving e^x


EDIT: I posted when you had written.... im reading your post now, thanks

 

[pka]

\(\displaystyle \L 5^x = 10\quad \Rightarrow \quad x = \frac{{\ln (10)}}{{\ln (5)}}\)

 

[stapel]

I don't understand how you would solve 5^x = 10 using ln.

You'd solve it in the same way you would using any other base log: Take the log of both sides, move the x down, and divide through.

. . . . .5^x = 10
. . . . .log₂(5^x) = log₂(10)
. . . . .x log₂(5) = log₂(10)
. . . . .x = log₂(10) / log₂(5)

. . . . .5^x = 10
. . . . .log₃(5^x) = log₃(10)
. . . . .x log₃(5) = log₃(10)
. . . . .x = log₃(10) / log₃(5)

. . . . .5^x = 10
. . . . .log₄(5^x) = log₄(10)
. . . . .x log₄(5) = log₄(10)
. . . . .x = log₄(10) / log₄(5)

. . . . .5^x = 10
. . . . .log_8.3(5^x) = log_8.3(10)
. . . . .x log_8.3(5) = log_8.3(10)
. . . . .x = log_8.3(10) / log_8.3(5)

. . . . .5^x = 10
. . . . .log_#(5^x) = log_#(10)
. . . . .x log_#(5) = log_#(10)
. . . . .x = log_#(10) / log_#(5)

. . . . .5^x = 10
. . . . .log_;-)(5^x) = log_;-)(10)
. . . . .x log_;-)(5) = log_;-)(10)
. . . . .x = log_;-)(10) / log_;-)(5)

. . . . .5^x = 10
. . . . .log₁₀(5^x) = log₁₀(10)
. . . . .x log₁₀(5) = log₁₀(10)
. . . . .x = log₁₀(10) / log₁₀(5)

. . . . .5^x = 10
. . . . .log_e(5^x) = log_e(10)
. . . . .x log_e(5) = log_e(10)
. . . . .x = log_e(10) / log_e(5)

Only the last two of these can be entered directly into one's calculator.

Eliz.

 

[jwpaine]

Thanks Stapel. I was looking for some type of explanation, such as the examples you gave.

I was confused on why you can use any base in the rational and still come out with the same answer.


Quoting from a website I found:

"It sounds like a pretty strange world with such a logarithm as that can be considered natural, but it turns out that this logarithm is the most useful one for calculus. In fact for many people by the time they get through calculus they are convinced that God must have used base e when he created the universe, because it comes up in so many places. But right now I don't have time to explain all of that, so you will have to take my word for it. Anyway, if you go on to much more advanced mathematics classes the only log that will be used is log base e."

I'm in algebra III and we didn't cover logs very long... only like a week or so..so I guess I'll understand eventually.

 

[tkhunny]

9^(5x-3)=78
why would you use anything but base 10 for this?

One more thing. Why would Base 10 be the most obvious for this probem? Frankly, Base 9 or Base 3 look like they might be a little more useful.

\(\displaystyle \L\;5x-3\;=\;\log_{9}(78)\)

or

\(\displaystyle \L\;9^{5x-3}\;=\;(3^{2})^{5x-3}\;=\;3^{10x-6}\)

\(\displaystyle \L\;10x-6\;=\;\log_{3}(78)\)

From a perspective of notation, they certainly are cleaner than 10 or e. Of course, I don't have a Base 9 or Base 3 button on my calculator, either.

If it were 76, rather than 78, it would be even cuter.
If it were 81, rather than 78, guess which method would win this argument?