# Thread: equations to log form, is this correct?

1. ## equations to log form, is this correct?

can someone tell me if i am doing this right?

Convert the following equations into logarithmic form: y=bx logb(y)=x

a. 9 = 4x 9=4^x log4(9)=x

b. 3 = 6y 3=6^y log6(3)=y

c. 5 = 7y 5=7^y log7(5)=y

d. X = 9y x=9^y log9(x)=y

2. Originally Posted by brandilee
can someone tell me if i am doing this right?

Convert the following equations into logarithmic form: y=bx logb(y)=x

a. 9 = 4x 9=4^x log4(9)=x

b. 3 = 6y 3=6^y log6(3)=y

c. 5 = 7y 5=7^y log7(5)=y

d. X = 9y x=9^y log9(x)=y
Whaaaat?
If y = bx then $\displaystyle x \neq log_b(y)$

If y = b^x then $\displaystyle x = log_b(y)$. Is this what you are trying to say? (This is correct.)

Then what's this "9 = 4x" thing in front of the "$\displaystyle 9 = 4^x$?"

-Dan

3. (I assume that you accidentally wrote those 9 = 4x, 3 = 6y, 5 = 7y..)

$\displaystyle 4^x = 9$,
$\displaystyle x = \text{log}_49$

$\displaystyle 6^y = 3$,
$\displaystyle y = \text{log}_63$

$\displaystyle 7^y = 5$,
$\displaystyle y = \text{log}_75$

$\displaystyle 9^y = x$,
$\displaystyle y = \text{log}_9x$

4. yeah, thats what i meant. i dont know how to make the numbers above or below. thanks