How do I make $x$ the subject when it is inside a log fraction?

Question

I have an equation:

$y=145366.45\left(1-\left(\frac{x}{1013.25}\right)^{0.190284}\right)$

$\frac{y}{145366.25}=1-\left(\frac{x}{1013.25}\right)^{0.190284}$

$\ln{\left(\frac{y}{145366.25}\right)}=1- 0.190284 \ln{\left(\frac{x}{1013.25}\right)}$

$\ln{\left(\frac{y}{145366.25}\right)} -1 = - 0.190284 \ln{\left(\frac{x}{1013.25}\right)}$

$\frac{\ln{\left(\frac{y}{145366.25}\right)}}{-0.190284} -1 = \ln{\left(\frac{x}{1013.25}\right)}$

What do I do with

$\ln{\left(\frac{x}{1013.25}\right)}$

so that I can make $x$ the subject?

Yes, I'm clearly a math noob.

Answer 1

Set $a=1013.25$, $b=0.190284$, $c=145366.45$, so your expression becomes $$ y=\left(1-\left(\frac{x}{a}\right)^{b}\right)c $$ Then $$ \frac{y}{c}=1-\left(\frac{x}{a}\right)^{b} $$ and therefore $$ \left(\frac{x}{a}\right)^{b}=1-\frac{y}{c} $$ so $$ \frac{x}{a}=\left(1-\frac{y}{c}\right)^{1/b} $$ and finally $$ x=a\left(1-\frac{y}{c}\right)^{1/b} $$

Answer 2

I will answer your question "...so that I can make $x$ the subject?", but first, you have an error.

In moving from line 2 to line 3, you apply the natural log to both sides. However, you cannot use $\ln$ the way you do.

I will change your equation to make the example easier to follow.

Line 2 is something like $\frac{y}{14}=1-x^{19}$.

When applying $\ln$ to both sides, we must apply it to both sides.

Line 3 should be like $\ln\left(\frac{y}{14}\right)=\ln\left(1-x^{19}\right)$. In your work, you didn't have the $1$ in the argument of $\ln$.

So, what to do? We need to isolate the $x$-term before using $\ln$.

Working from Line 2...

\begin{align} \frac{y}{145366.25}&=1-\left(\frac{x}{1013.25}\right)^{0.190284}\\ \frac{y}{145366.25}-1&=-\left(\frac{x}{1013.25}\right)^{0.190284}&\text{Isolate $x$-term}\\ 1-\frac{y}{145366.25}&=\left(\frac{x}{1013.25}\right)^{0.190284}&\text{Divide by $-1$}\\ \ln\left(1-\frac{y}{145366.25}\right)&=\ln\left[\left(\frac{x}{1013.25}\right)^{0.190284}\right]&\text{Apply $\ln$ to both sides}\\ \ln\left(1-\frac{y}{145366.25}\right)&=0.190284\ln\left(\frac{x}{1013.25}\right)&\text{Power Rule for $\ln$}\\ \frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln\left(\frac{x}{1013.25}\right) \end{align}

Now, what to do with $\ln\left(\frac{x}{1013.25}\right)$? We use the Quotient Rule for Logarithms. $$\ln\left(\frac{a}{b}\right)=\ln a-\ln b$$

Picking up where we stopped...

\begin{align} \frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln\left(\frac{x}{1013.25}\right)\\ \frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln x -\ln 1013.25&\text{Quotient Rule for Logs}\\ \frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}+\ln 1013.25&=\ln x \end{align}

Raise both sides to base $e$, or use the inverse property of the natural log, and you have $x$ isolated.