Question bank — Graphs of logarithmic functions

False. Reflecting $y=a^x$ in $y=x$ turns its horizontal asymptote $y=0$ into a vertical one $x=0$; the curve dives to $-\infty$, it never levels off.

True. ${\log }_{a}1=0$ for every base because $a^0=1$, so the $x$-intercept is fixed at $(1,0)$ regardless of $a$.

False. The $y$-axis is $x=0$, which is the vertical asymptote and excluded from the domain, so there is no $y$-intercept at all.

True in shape/features (both increasing, same asymptote and $(1,0)$), but they differ in steepness — they only agree at $(1,0)$ and cross-scale by the constant factor $\ln 10$.

True. Since ${\log }_{a}x=-{\log }_{1/a}x$ and $1/a>1$, it's the increasing curve flipped in the $x$-axis, so it falls.

True. Its range is all reals because the exponential $y=a^x$ has domain all reals, and reflecting swaps domain and range.

False. The $+2$ is inside the function, so it acts on $x$ (horizontal): the graph shifts left 2, and the asymptote moves to $x=-2$.

True. ${\log }_{1/a}x=-{\log }_{a}x$, so both are the reflection of $y={\log }_{a}x$ in the $x$-axis.

Error: $x=a^y$ is always positive, so $\log$ of $0$ or a negative number doesn't exist. Domain is strictly $x>0$.

Error: it does — logs are unbounded above. Set ${\log }_{2}x=3\Rightarrow x=2^3=8$, so it passes through $(8,3)$; it rises forever, just slowly.

Error: by log laws ${\log }_a(kx)={\log }_{a}k+{\log }_{a}x$, so multiplying inside is identical to a vertical shift up by ${\log }_{a}k$ — same shape, moved up/down.

Error: the $-3$ shifts the whole curve right 3, dragging the asymptote to $x=3$. Domain becomes $x>3$.

Error: as inverse functions they're reflections in $y=x$, but for $a>1$ they may not intersect at all (e.g. $a=10$); intersections needn't lie on $y=x$ even when they exist.

Error: $y=0$ needs ${\log }_a(x-b)=0$, i.e. $x-b=1$, so $x=b+1$. The intercept moves with the horizontal shift.

Error: bigger base makes it rise slower. A larger $a$ means you must multiply $x$ by a bigger factor to gain each unit of $y$, so the curve is flatter.

Because ${\log }_{a}a=1$: rewriting as $a^1=a$ shows the input $a$ gives output $1$, an anchor point on every log curve.

To raise $y$ by 1 you must multiply $x$ by $a$. Each extra unit of height costs an ever-larger horizontal jump, so the slope flattens.

Because $y=a^x$ has asymptote $y=0$; reflecting in $y=x$ swaps this to $x=0$. The exponential never outputs a negative or zero value.

The log law ${\log }_a(kx)={\log }_{a}k+{\log }_{a}x$ converts the inside-multiplication into an added constant ${\log }_{a}k$, which is literally a vertical translation.

Because a multiplicative jump in the input becomes an additive step in the output; a factor of $1000000$ in $x$ only adds 6 to ${\log }_{10}x$, squeezing enormous ranges into small numbers.

Because $a=1$ gives $1^y=1$ always (no unique inverse), and $a\le 0$ makes $a^y$ undefined for many $y$ — so no well-defined reflection to invert.

$y\to -\infty$. Tiny positive $x$ needs a very large negative power of $a$, so the curve plunges down the asymptote $x=0$.

Undefined — $x=0$ is not in the domain; you cannot take the log of zero, so there is a "hole" (asymptote), not a point.

No. For $a>1$ it's strictly increasing forever; for $0<a<1$ strictly decreasing — monotone throughout, so no max, min, or turning point.

We need $3-x>0$, so $x<3$. The reflection in the $y$-axis (from the $-x$) flips the graph and puts the asymptote at $x=3$, valid only to its left.

At $x=e$, $y=1$; at $x=1/e$, $y=-1$ (since $e^{-1}=1/e$). These sit symmetrically about the $x$-intercept $(1,0)$ in the log's own scale.

Yes — $x=1$ is in the domain (positive) and gives $y=0$, the guaranteed $x$-intercept for every base.