1. ## this should be really simple (Headbang)

2^y+1 = 3x 5^y

2. Originally Posted by scouser
2^y+1 = 3x 5^y
I think the above means the following:

. . . . . $2^y\, +\, 1\, =\, (3x)(5^y)$

But what are you supposed to do with this? What were the instructions?

Thank you!

3. ## sorry

ok
2 to the power of y+1 is equal to 3 times (5 to the power of y)
Im supposed to find y

4. Originally Posted by scouser
2 to the power of y+1 is equal to 3 times (5 to the power of y)
Im supposed to find y
So you are supposed to solve the equation, which is as follows:

. . . . . $2^{y+1}\, =\, 3\left(5^y\right)$

A good start for solving this exponential equation would probably be to take the log of either side:

. . . . . $\ln(2^{y+1})\, =\, \ln(3(5^y))$

Apply a log rule to split apart the right-hand side:

. . . . . $\ln(2^{y+1})\, =\, \ln(3)\, +\, \ln(5^y)$

Use the definition of logs to simplify as:

. . . . . $(y\, +\, 1)\ln(2)\, =\, \ln(3)\, +\, (y)\ln(5)$

Then solve this linear equation for "y=".

5. ## and then

then im stuck!
are you stuck also hahahahaha?
joking

6. ln 5, ln2 and ln 3 are just numbers!

Solve it for y as such!

(I can also safely say that Stapel isn't stuck!)