# Thread: Solution to Exponential equation

1. ## Solution to Exponential equation

Can this equation be solved explicitly for x
$y=2.3^x + 7^x$

2. ## Re: Solution to Exponential equation

I think there're no real solutions if you want to solve y=0.

3. ## Re: Solution to Exponential equation

Originally Posted by Siron
I think there're no real solutions if you want to solve y=0.
That is not the question (but you can see that y cannot be zero or negative starting in two lines...)

$y = a^x + b^x$

$\ln y = x \ln a + x \ln b$

$\ln y = x (\ln a + \ln b)$

$\ln y = x (\ln ab)$

Can you take it from here?

4. ## Re: Solution to Exponential equation

If you are given a y solve for x, I am looking for an algorithm

5. ## Re: Solution to Exponential equation

Originally Posted by TheChaz
That is not the question.

$y = a^x + b^x$

$\ln y = x \ln a + x \ln b$

$\ln y = x (\ln a + \ln b)$

$\ln y = x (\ln ab)$

Can you take it from here?
Someone needs their morning coffee.

Let $x=2$, $a=3$, $b=4$

6. ## Re: Solution to Exponential equation

Originally Posted by TheChaz
That is not the question (but you can see that y cannot be zero or negative starting in two lines...)

$y = a^x + b^x$

$\ln y = x \ln a + x \ln b$

$\ln y = x (\ln a + \ln b)$

$\ln y = x (\ln ab)$

Can you take it from here?
Except $\ln(a^x+b^x) \neq x\ln(a) + x\ln(b)$

7. ## Re: Solution to Exponential equation

Originally Posted by Quacky
Someone needs their morning coffee.

Let $x=2$, $a=3$, $b=4$
I must need some coffee... Then y = 25 ??
Can you come to the punchline - I'm a little tired apparently!

8. ## Re: Solution to Exponential equation

Originally Posted by e^(i*pi)
Except $\ln(a^x+b^x) \neq x\ln(a) + x\ln(b)$
Oh *^&%& ! Hah... I knew it seemed too easy!

(edit... I did have a pretty good night last night )