Simple exponential equations

Question

Good evening to everybody, I have a doubt about the following two equations:

A. $2^{x+1}=5^{1-x}$

B. $3^x+2*3^{1-x}=29/3$

I know that x should equal to 9

.

How do I arrive to this result? ex Thanks in advance!

Answer 1

For the first equation, $2^{x+1}=5^{1-x}$

$x+1=(1-x)log_2(5)$

$x(1+log_2(5))=log_2(5)$

So $x=\frac{log_2(5)}{1+log_2(5)}$. I'm not sure what you meant about x being 9, since that definitely doesn't work. I'm assuming your two equations are seperate, so now I will solve the second one:

$3^x+6*3^{-x}=29/3$

Letting $y=3^{x}$, we have $y+\frac{6}{y}=29/3$, or $3y^2-29y+18=0$, which, with the quadratic formula, gives

$y=\frac{29 \pm 25}{6}$, so the two solutions are 2/3 and 9, which means that $x=2$ or $x=log_3(2)-1$, which are different to in part a