# Thread: this should be really simple (Headbang)

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

sorry i expressed that badly
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

yeah i had got up to there already.
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!)

Read the links that have been posted and post back your attempt.

7. Originally Posted by scouser
yeah i had got up to there already.
then im stuck!
In future, please show the work you've done.

In this case, I showed the hard part on the assumption that you'd already studied the easy part. (It's highly unusual to teach how to work with logarithms before having taught how to solve linear equations.) To learn how to do the remaining step, please try here.

8. oh sure, i get it
thanks