How do I solve for x on this problem: log(x+4)-log(x+1)=1

Answer 1

First, use the quotient rule for logarithms to rearrange the left side of the equation. Here is the rule:

log_b(x/y) = log_b(x) - log_b(y)

Using this rule, we can rewrite log(x+4) - log(x+1) as

log((x+4)/(x+1))

So, now we have log((x+4)/(x+1)) = 1

We can now rewrite the equation again, using the inverse function rule:

If log_b(x) = y, then b^y = x

Since this is precalculus, we can assume a base of 10 for logarithm, since it does not specify in the original equation. Therefore, our 'b' from above will be 10. We then have:

10¹ = (x+4)/(x+1)

10 = (x+4)/(x+1)

10(x+1) = (x+4)

10x + 10 = x+4

9x= -6

x=-6/9

x=-2/3