# Thread: solving for y in terms of x

1. ## solving for y in terms of x

i dont know how to go about this problem, solve the following for y in terms of x: 10^y=5^x-3
thank you!

2. Originally Posted by mxhockey140
i dont know how to go about this problem, solve the following for y in terms of x: 10^y=5^x-3
thank you!
Before I proceed, is it $\displaystyle 10^y=5^{x-3}$ or $\displaystyle 10^y=5^x-3$?

3. its the first one x-3 is the exponent

4. Originally Posted by mxhockey140
its the first one x-3 is the exponent
The way to do this involves taking the natural log of both sides:

$\displaystyle 10^y=5^{x-3}\implies \ln\left(10^y\right)=\ln\left(5^{x-3}\right)$

Using power rules, we can rewrite this as $\displaystyle y\ln\left(10\right)=\left(x-3\right)\ln\left(5\right)$

Can you take it from here?

5. no not really.. im not the best with this stuff.

6. Originally Posted by mxhockey140
no not really.. im not the best with this stuff.
If you're attempting questions involving exponentials and logarithms it's expected that you can make y the subject by solving what is essentially a linear equation.

Can you solve $\displaystyle a y = b (x - 3)$ for y?

7. thats why i was looking for help..

8. Originally Posted by mxhockey140
thats why i was looking for help..
I would have assumed that you were looking for help in getting from the original equation to a stage where you could reasonably be expected to complete the solution.

Since another question you've asked (http://www.mathhelpforum.com/math-he...28-curves.html) strongly suggest that you're studying calculus, your trouble in solving a linear equation is surprising and of concern.

Can you solve for y?

What about solving $\displaystyle 3y = 4(x - 3)$ for y?

If you cannot at least do this last one then I would suggest that your real problem is an insufficient grasp of the mathematical knowledge prerequisite to the topic you're currently being taught. In which case you're strongly advised to go back and revise that material in collaboration with your instructor.