Question

How do I solve this equation for x? (Will give medal & fan)

log(5x) + log(x - 1) = 2

Answer 1

raising a log to be the exponent of its base gives you whats in the brackets

so 10^(log(5x)) = 5x

you have to raise each thing as a power of 10 to make it balance

log(5x) is saying, "what number do i have to raise 10 to in order to get 5x"

10^log(5x) says "10 raised to the power that i have to raise 10 to in order to get 5x" (so you get 5x)

Answer 2

So wait, would I do the same thing for log(x - 1)? So 10^(log(x-1)) = x - 1 ?

Answer 3

exactly
but to make it balance you have to do the same thing on the other side

10^2

Answer 4

oh okay (: So

5x + x - 1 = 100
6x - 1 = 100
6x = 101
x = 101/6
x = 16.833

Answer 5

Is that right?

Answer 6

Not correct. You have to change the problem to the standard form before you try and use the equivalency. Like this\[\log(5x) + \log(x - 1) = 2 \implies \log(5x^2-5x)=2 \implies 5x^2-5x=100\]Now you have a quadratic equation you can solve. When you get solutions, make sure your answer is in the domain of the original problem.

Answer 7

alright sorry

Answer 8

The equivalency is this one (very important for these kinds of problems):\[y=a^x \iff \log_a y=x\]

Answer 9

To finish the problem, you need to do something similar to this:\[5x^2-5x-100=0 \implies x^2-x-20=0 \implies (x-5)(x+4)=0\]\[\implies x=5~or~x=-4\]The second solution isn't in the domain of the original problem, so the solution is x=5.