# Thread: help with a few problems (don't need to answer all)

1. ## help with a few problems (don't need to answer all)

i have tried these many times and still can't get them.

1) f(x)= ((5x^3)-(5x^2)+9x-9) / (x-1) when x < 1
f(x)= (6x^2)-1x+a when x>=1

What value is a to make the function continuous at 1?

I tried making them equal to each other and putting x=1 but that didn't work.

2) (e^2x) - (7e^x) + 6 = 0

It says there are 2 solutions but I keep getting -33.1255 and that's not even one of them.

3) y=loga(x) with the subscript being a. It passes through (28,1).

What is the value of a?

I have no idea what to do on this one.

4) log2(x) + log2(x+3) = k

Solve for x in terms of k.

Again, I have no idea what to do. I can solve it if k is an integer but thats about it.

Any help is greatly appreciated.

2. 1. Your approach would normally work, except that the function $\frac{5x^3 - 5x^2 + 9x - 9}{x - 1}$ has a discontinuity at x = 1. But what is the limit of the function? Factor the numerator to yield $\frac{(5x^2 + 9)(x-1)}{x-1}$. Then cancel the factors of $x-1$ and you are left with $5x^2 + 9$. At x = 1, this function is equal to 14, so the limit is 14, and that is the function value you want for the other part of your function.

2. Substitute $u = e^x$, yielding $u^2 - 7u + 6 = 0$. Then solve for u by factoring and identifying the roots: $(u - 6)(u - 1) = 0$, so u = 1 or 6. Hence $e^x = 1$ or $e^x = 6$. From $e^x = 1$ you have the solution x = 0, and from $e^x = 6$ you have the solution x = ln 6.

3. This is an easy one. The equation $y = \log_a x$ can be rewritten as $a^y = x$. With (28, 1) on the graph, we have $a^1 = 28$, or $a = 28$.

4. You can combine the two logarithms to yield $\log_2 (x(x+3)) = k$, which can be written $2^k = x(x + 3)$, or $x^2 + 3x - 2^k = 0$. Now apply the quadratic formula: $x = \frac{-3 \pm \sqrt{9 + 4\cdot 2^k}}{2}$. Recognizing that x must be a positive number in order to evaluate the logarithm, we have $x = \frac{-3 + \sqrt{9 + 2^{k+2}}}{2}$.

3. ooooooo. Those explanations were great. Thanks a lot.