Solution to Exponential equation

**Goku** · August 7th 2011, 06:38 AM

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

**Siron** · August 7th 2011, 07:11 AM

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

**TheChaz** · August 7th 2011, 07:46 AM

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?

**Goku** · August 7th 2011, 07:48 AM

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

**Quacky** · August 7th 2011, 07:50 AM

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

**e^(i*pi)** · August 7th 2011, 07:57 AM

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 $\ln(a^x+b^x) \neq x\ln(a) + x\ln(b)$

**TheChaz** · August 7th 2011, 08:00 AM

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!

**TheChaz** · August 7th 2011, 08:02 AM

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

)

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