Answer: (a) \(x=-\sqrt[3]{3}\). (b) The graph has \(y\)-intercept \((0,7)\), no \(x\)-intercept, and horizontal asymptote \(y=3\).
Answer: (a) \(x=-\sqrt[3]{3}\). (b) The graph has \(y\)-intercept \((0,7)\), no \(x\)-intercept, and horizontal asymptote \(y=3\).
(a) Write all terms as powers of 5:
\(625=5^4\) and \(125=5^3\).
So
\(\displaystyle \frac{625^{(x^3-1)/2}}{125^{x^3}}=\frac{(5^4)^{(x^3-1)/2}}{(5^3)^{x^3}}\).
This is
\(\displaystyle \frac{5^{2x^3-2}}{5^{3x^3}}=5^{2x^3-2-3x^3}=5^{-x^3-2}\).
The equation becomes
\(5^{-x^3-2}=5^1\).
Equating powers gives
\(-x^3-2=1\),
so
\(x^3=-3\).
Therefore
\(x=-\sqrt[3]{3}\).
(b) For \(y=4\mathrm{e}^x+3\), when \(x=0\),
\(y=4\mathrm{e}^0+3=7\).
So the \(y\)-intercept is \((0,7)\).
Since \(\mathrm{e}^x\gt 0\) for all real \(x\), \(4\mathrm{e}^x+3\gt 3\), so there is no \(x\)-intercept.
As \(x\to-\infty\), \(\mathrm{e}^x\to0\), so the horizontal asymptote is
\(y=3\).
