Solving Exponential Equations? Help?

Question

Hello, I am having some trouble with exponential equations here. 

 

2^1-3x = 5^x+1

 

The problem I am having with this equation here, is that there is a 3x, which is locked together, and the only way you can do that is divide, right? 

 

I guess my question is, how would i divide out the 3 from the x, or the x from the 3, so I can go ahead and solve my way to X? 

 

Thanks again.

 

Answer 1

So, I think your equation looks like this:

 

2^1-3x = 5^x+1,

 

where 1-3x and x+1 are both entirely in the exponents. Right?

 

 

If that's the case, start out by taking the log of both sides:

 

log(2^1-3x) = log(5^x+1)

 

Remember that log(aⁿ) = nlog(a). This lets us take those x's in the exponents out of the exponents:

 

(1-3x)log2 = (x+1)log5

 

But, it's still hard to see how you can detach that x from the 3. To do so, we'll first distribute both sides:

 

log2 - 3xlog2 = xlog5 + log5

 

Now, get everything with an x in it onto one side and everything else onto the other side:

 

log2 - log5 = xlog5 + 3xlog2

 

Finally, factor the x out of the right-hand side and divide by whatever's left to isolate x:

 

log2 - log5 = x(log5 + 3log2)

(log2 - log5)/(log5 + 3log2) = x

 

That means x = (log2 - log5)/(log5 + 3log2). You can either just plug those numbers in or simplify first to x = log(2/5)/log(5×2³) = log(2/5)/log(40). Both are approximately equal to -0.2483. So, that's your value for x.

 

 

The trick here is that, once you've gotten the x's out of the exponents using logs, you should move all the terms with x in them onto one side by themselves. Then, you can factor the x out and divide by what remains.