Looking for clarification on this equation: solve y=-12.34ln(x)+133.72 for x

[JM4]

Hi there,

I have this equation:

y=-12.34ln(x)+133.72

Can someone please clarify how to solve for x?

Pretty standard stuff, but I haven't done much albegra or solving for x/y in a loonngg time.

cheers.

 

[mmm4444bot]

Looking for clarification on this equation …

… I haven't done much albegra or solving for x/y in a loonngg time …

Google keywords solving logarithmic equations, to find lots of video lectures or written lessons and examples. Once you've refreshed your memory, try your exercise again.

If you get stuck, please show us what you tried or explain why you're stuck. Of course, if you see something in a lesson that you don't understand, ask us about that, first.

Also, read the forum guidelines; you may start with this summary. Thank you! :cool:

 

[JeffM]

Hi there,

I have this equation:

y=-12.34ln(x)+133.72

Can someone please clarify how to solve for x?

Pretty standard stuff, but I haven't done much albegra or solving for x/y in a loonngg time.

cheers.

I am presuming that this not a homework problem (it's too messy).

You cannot get a numeric answer to this equation. In general, if you have n unknowns, you need n equations to get numeric answers.

If all you want is to re-aarange this equation so x is defined in terms of y rather than y being defined in terms of x:

\(\displaystyle y = 12.34ln(x) + 133.72 \implies \\

12.34ln(x) = y - 133.72 \implies \\

ln(x) = \dfrac{y - 133.72}{12.34} \implies \\

x = e^{\{(y - 133.72) \div 12.34\}}.\)

 

[JM4]

Sorry, I was not clear at all! using this equation to solve for y (i.e have x values).

Cheers!

 

[Deleted member 4993]

Sorry, I was not clear at all! using this equation to solve for y (i.e have x values).

Cheers!

Are you allowed to use a scientific calculator (or software like MS_Excel)?

 

[mmm4444bot]

… using this equation to solve for y …

i.e have x values …

In post #1, you asked how to solve for x; now you say you know x, and you're asking how to solve for y.

I don't understand because the equation you posted is already solved for y.

(Maybe you're typing 'solve', while thinking 'evaluate'.)

Given a specific value of x, you can evaluate the given expression for y by substitution and using software (eg: scientific calculator) to approximate the natural log.

EG: x = 8.95

y = -12.34 ln(x) + 133.72

y = -12.34 ln(8.95) + 133.72

Software says: ln(8.95) ≈ 2.1917

y ≈ -12.34(2.1917) + 133.72

y ≈ 106.67 (rounded)

 

[mmm4444bot]

… \(\displaystyle y = 12.34 ln(x) + 133.72\) …

The negative sign on 12.34 is missing. (Easy to overlook, when people don't insert spaces around their equal signs). :cool:

 

[pka]

Hi there,
I have this equation:
y=-12.34ln(x)+133.72
Can someone please clarify how to solve for x?
Pretty standard stuff, but I haven't done much albegra or solving for x/y in a loonngg time.

Notation to make matters easy: \(\displaystyle \exp(f(x))=e^{f(x)}\)


\(\displaystyle \begin{align*}y&=-12.34\log(x)+133.72\\12.34\log(x)&=133.72-y\\\log(x)&=\dfrac{133.72-y}{12.34}\\x&=\exp\left(\dfrac{133.72-y}{12.34}\right)\end{align*}\)