Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to remove exponentiation in this equation $y=x^t$ and use only multiplication or division. I have x, t and ln(x). Is it possible?

share|improve this question
    
$\ln y=t\ln x$ is the best you can do. –  David Mitra Feb 6 '13 at 14:56
add comment

2 Answers

up vote 1 down vote accepted

$y=x^t$

This gives you: $\ln(y) = \ln(x^t) = t \ln(x)\quad\implies \quad \ln(y) = t \ln(x)$

Here we use the rule of logarithms (which applies to "$\ln$" as well):

$$\log(a^b) = b \log (a)$$

share|improve this answer
    
So now instead using pow I have to use exp. What if I could also use addition and subtraction? –  Martin Feb 6 '13 at 15:25
    
That's about as simplified as you can get. You could write: $t = \dfrac{\ln(y)}{\ln(x)}$ –  amWhy Feb 6 '13 at 15:32
    
To solve for $y$ given $t$ and $\ln(x)$, evaluate $t\ln(x) = a$, so $\ln(y) = a \implies y = e^a$. I.e. $y = \exp(t\ln(x))$ –  amWhy Feb 6 '13 at 15:38
add comment

When $a>0$ then we are always allowed to have $$\ln(a^b)=b\ln(a)$$ So take a Neperian logarithm from both sides of your identity.

share|improve this answer
    
What is the Neperian logarithm? Natural log? +1 –  amWhy Feb 8 '13 at 0:07
    
yes. it is the natural one. –  B. S. Feb 8 '13 at 5:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.