Getting rid off the logarithms in an equation to simplify

$\begingroup$

ok, I'm having trouble solving for equations when logarithms are involved. I know a little bit about logarithm rules but in equations I'm lost. example:

$$\frac{1}{b}\ln{y}=\frac{1}{a}\ln{x}+c$$

I can take $b$ to the other side: $$\ln{y}=\frac{b}{a}\ln{x}+bc$$

I know you can solve for $y$ by incorporating $e$ to both sides of the equation. and there's where I have trouble with.

How you add $e$ to the right side ?

$\endgroup$

2 Answers

$\begingroup$

$$\ln{y}=\frac{b}{a}\ln{x}+bc$$ $$\ln{y}=\frac{b}{a}\ln{x}+\ln(e^{bc})$$ $$\ln{y}=\ln{x^{\frac{b}{a}}}+\ln(e^{bc})$$ $$\ln{y}=\ln({x^{\frac{b}{a}}e^{bc}})$$ $${y}={x^{\frac{b}{a}}e^{bc}}$$

$\endgroup$ 1 $\begingroup$

$$\ln y = \frac{b}{a}\ln x + bc$$

Note: $\ln p = q \implies p = e^q$

$$ y = e^{\frac{b}{a}\ln x + bc} = e^{\ln x^{\frac{b}{a}}}\cdot e^{bc}$$

$$ y = x^\frac{b}{a}e^{bc}$$

Note: $e^{\ln m} = m$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy