Question

2^x=3^x+1 what are the exact approximate solutions

Answer 1

Answer:

d.

Step-by-step explanation:

The goal of course is to solve for x.  Right now there are 2 of them, one on each side of the equals sign, and they are both in exponential positions.  We have to get them out of that position.  The way we do that is by taking the natural log of both sides.  The power rule then says we can move the exponents down in front.

$ln(2^x)=ln(3^{x+1})$ becomes, after following the power rule:

x ln(2) = (x + 1) ln(3).  We will distribute on the right side to get

x ln(2) = x ln(3) + 1 ln(3).  The goal is to solve for x, so we will get both of them on the same side:

x ln(2) - x ln(3) = ln(3).  We can now factor out the common x on the left to get:

x(ln2 - ln3) = ln3.  The rule that "undoes" that division is the quotient rule backwards.  Before that was a subtraction problem it was a division, so we put it back that way and get:

$x(\frac{ln(2)}{ln(3)})=ln(3)$.  We can factor out the ln from the left to simplify a bit:

$x[ln(\frac{2}{3})]=ln(3)$.  Divide both sides by ln(2/3) to get the x all alone:

$x=\frac{ln(3)}{ln(\frac{2}{3}) }$

On your calculator, you will find that this is approximately -2.709