algebra-precalculuslogarithms
I am to solve $\ln5+\ln(5x^2-5)=\ln56$ using the one to one property of logs. According to my textbook the solution is $\pm\frac{9}{5}$ whereas I got $\sqrt{\frac{56}{5}}$
My working:
$$\ln5+\ln(5x^2-5)=\ln56$$
$$5x^2-5+5=56$$
$$5x^2=56$$
$$x^2=\frac{56}{5}$$
$$x=\sqrt{\frac{56}{5}}$$
$$~3.35$$
Where did I go wrong and how can I arrive at $\pm\frac{9}{5}$?
The root is $x = -4$. So you have to take $-4$ instead of $4$
Hence you have $2x^2-2x-3$
Notice how the graph has a horizontal asymptote at $y = 7$. This is from the graph being shifted up by 7. This means we should use the form:
$y = ab^x + c$, where c is the horizontal asymptote. This form has technically been used by you before, it's just that c normally equals $0$ so there's no need to write it. E.g., if you graph $y = 2^x$, you'll notice the horizontal asymptote is at $y = 0$ (so the graph hasn't been shifted).
So we have $y = ab^x + 7$. Now for this, because the graph has been shifted you can't get $a$ immediately just by looking at the graph's y-intercept. Instead you need to plug in the y-intercept's x and y coordinates in order to solve for $a$:
$5 = ab^0 + 7$
$5 = a(1) + 7$
$a = -2$
Notice how the $b$ was eliminated above since it was raised to $0$. This is thanks to the x-coordinate of the y-intercept being $0$. If we instead tried to use the other point on the graph first, you'd be left with both $a$ and $b$ in the equation.
Anyway, now that we've solved for $a$, plug -2 into the equation:
$y = -2b^x + 7$
There's only one variable left to solve: $b$. So use the (1,1) coordinate on the graph:
$1 = -2b^1 + 7$
$-6 = -2b$
$b = 3$
So we get: $y = -2(3)^x + 7$
Note - the following video explains a similar problem to yours. The graph is flipped over the y-axis but the method is the same.
Best Answer
$\ln a+\ln b=\ln c$ does not imply $a+b=c$, i.e. $\ln$ does not distribute over addition. Instead it implies $ab=c$: $$5(5x^2-5)=56$$ $$x^2-1=\frac{56}{25}$$ $$x^2=\frac{81}{25}\implies x=\pm\frac95$$