# Thread: Exponential and logarithmic equations

1. ## Exponential and logarithmic equations

I was having trouble with the following:

1)An expression for y in terms of x by rearranging 18loge(x)=6-3loge(y).

2) solutions for x in (log6(x))^2+4log3(6)+6=0.
-Do i interpret this one as a quadratic equation? If so, the solutions should be 3 and 2, but from there what do i do?

-thanks.

2. 1. Rearrange: 3ln(y) = 6-18ln(x) = 6ln(e) - 18ln(x)

Now what do you know about your log laws that will allow you to simplify the RHS?

2. If they're the same base yes. If not you may need to use the change of base rule

3. Not really sure what they are after here, but is this correct?

18loge(x)=6-3loge(y)

restated as: 18*ln(x) = 6 - 3*ln(y)

express in terms of y=

18*ln(x) + 3*ln(y) = 6

ln(x^18y^3) = 6

e^6 = x^18y^3

y^3 = e^6/x^18

y = cube root(e^6/x^18)

4. Hi angypangy,
Yout last equation can be firther simplified.

bjh

5. You mean:

y = e^2
-----
x^6

?

6. You may rewrite your last equation as: $\sqrt[3]{\left(\dfrac{e^2}{x^6}\right)^3}$

Obviously you can simplify that further. In addition we need to state that $x,y > 0$ - do you see why?

Originally Posted by angypangy
You mean:

y = e^16/3
-----
x^6

?
Where did 16 come from? The line before has e^6 (which is correct) so the exponent on e is 6/2 = 3

7. Why can you not just state:
y = e^2
-----
x^6

8. You can and it is correct.

Nb: If you're going to write in plain text please use "/" for division. In other words y = (e^2)/(x^6)