Exponent problem

[displacerface]

I'm not entirely sure which math category this question belongs in... I'm doing some math for a science worksheet, and have set up an equation, and i need to solve for x.

2^x=⁴/₃

I tried using mathway, but since I'm not a member it wouldn't let me see how they reached the solution of x=2-ln(3)/ln(2)... please show me how this works...

 

[stapel]

i need to solve for x.

2^x=⁴/₃

please show me how this works...

We really can't teach courses here, certainly not in a simple forum posting. We can, however, provide you with links to lessons online, from which you can study at your leisure.

Have you ever heard of logarithms before, or do you need lessons that start with that topic (for instance, here)? Thank you!

 

[Deleted member 4993]

I'm not entirely sure which math category this question belongs in... I'm doing some math for a science worksheet, and have set up an equation, and i need to solve for x.

2^x=⁴/₃

I tried using mathway, but since I'm not a member it wouldn't let me see how they reached the solution of x=2-ln(3)/ln(2)... please show me how this works...

Apply ln to both sides - and simplify.

 

[Quaid]

how [have] they reached the solution of x = 2 - ln(3)/ln(2)

Hi face:

They could have used some definitions and properties from arithmetic and algebra.


Arithmetic Property

(F - G)/H = F/H - G/H


Difference of Logarithms Property

ln(A) - ln(C) = ln(A/C)


Logarithm Change-of-Base Property

log_b(A) = ln(A)/ln(b)


Definition of Logarithm

log_b(A) = x means that b^x = A


I hope that these help you.

Cheers :cool:

PS: What's the urgency?

 

[displacerface]

Well that is not the correct solution.

Getting the correct solution requires using logarithms.

A law of logs is that \(\displaystyle a = b > 0 \iff log_c(a) = log_c(b).\)

Another law of logs is \(\displaystyle log_c(a^d) = d * log_c(a).\)

A third law of logs is \(\displaystyle log_c\left(\dfrac{e}{f}\right) = log_c(e) - log_c(f).\)

The combination ln just refers to a log with a base equal to the transcendental number e.

Putting those together you get

\(\displaystyle 2^x = \dfrac{4}{3} \implies ln(2^x) = ln\left(\dfrac{4}{3}\right) \implies x * ln(2) = ln(4) - ln(3) \implies x = \dfrac{ln(4) - ln(3)}{ln(2)}.\)

I noticed that, too, and was wondering if ln(4)/ln(2)=2 or not... Also, why must I have used ln and not log?

 

[displacerface]

Hi face:

They could have used some definitions and properties from arithmetic and algebra.


Arithmetic Property

(F - G)/H = F/H - G/H


Difference of Logarithms Property

ln(A) - ln(C) = ln(A/C)


Logarithm Change-of-Base Property

log_b(A) = ln(A)/ln(b)


Definition of Logarithm

log_b(A) = x means that b^x = A


I hope that these help you.

Cheers :cool:

PS: What's the urgency?

The urgency was that this was due the next day. Any way I could change the title of this thread now?

 

[Quaid]

Well that is not the correct solution.

Jeff, Jeff, Jeff … so quick on the draw!

A decimal approximation for 2 - ln(3)/ln(2) is 0.415037 (rounded).

What approximation do you get, for your result?

(By the way, applying my list in reverse is sufficient to answer the question quoted in my first reply. Try it!)

Cheers :cool:

 

[Quaid]

I noticed that, too

How's that?

was wondering if ln(4)/ln(2)=2 or not

Let me google that for you.

why must I have used ln and not log

You don't. You may use the Natural logarithm or log₁₀ or some other base, as long as you don't have specific instructions to the contrary.
(Be careful of context, however. Sometimes the symbol log means exactly the same thing as the symbol ln.)

:cool:

 

[markraz]

I'm not entirely sure which math category this question belongs in... I'm doing some math for a science worksheet, and have set up an equation, and i need to solve for x.

2^x=⁴/₃

I tried using mathway, but since I'm not a member it wouldn't let me see how they reached the solution of x=2-ln(3)/ln(2)... please show me how this works...

my casio fx-5800p solves this problem
check it out great calc.... it's default form is fractions