What is the Solution for Solving Exponential Equations with Unknown Variables?

[NZS92]

Solve for x,
(73/10)^(x-1)+5 - 3^(x+1)=0

 

[jedishrfu]

we can't solve this for you, please show your attempts at solving it first so we can comment on where you're stuck or got it worng.

 

[NZS92]

I literally have no idea, tried log, ln, e^x,

my first work:
7.3^(x-1)+5=3^(x+1)

log_(3) (7.3^(x-1)+5)=x+1

e^(log_3 (7.3^(x-1)+5) / e^x = e

(7.3^(x-1)+5)/e^x =e
7.3^(x-1)+5= ee^x
7.3^(x-1) -ee^x=-5

my second work,
7.3^(x-1)=3^(x+1)-5

(x-1)log(73/10)=log(3^(x+1)-5)

xlog(73/10)-log(73/10) = log(3^(x+1)-5)

(e^(xlog(73/10)) / (e^(log73/10) = 3^(x+1)-5

e^(xlog73/10) = 3^(x+1) e^(log73/10)-5e^(log73/10)

 

[Mentallic]

Just to clarify, you mean
$$\left(\frac{73}{10}\right)^{x-1}-3^{x+1}+5=0$$

right? Because I don't believe this expression can be solved algebraically. You can, of course, find a numerical solution to the problem however.

 

[jedishrfu]

I agree with Mentallic, if its as he wrote there is no simple algebraic solution only a numeric one.