If log₁₀3 = 0.4771 and log₁₀e = 0.4343, then what is the value of log₁₀30.5?

The correct answer is 1.484.

To find the value of ${\log }_{10}{3}^{0.5}$, we can use the properties of logarithms.

First, recall the power rule of logarithms, which states that for any base $b$ and numbers $x$ and $y$:
${\log }_b\left(x^y\right)=y·{\log }_b\left(x\right)$

Applying this power rule to our expression, we can bring the exponent $0.5$ to the front of the logarithm:
${\log }_{10}{3}^{0.5}=0.5·{\log }_{10}3$

Now, we substitute the given value of ${\log }_{10}3=0.4771$ into the equation:
$0.5·0.4771=0.23855$

However, let us re-examine the expression in the question: "log₁₀30.5". This can be interpreted in two ways: either ${\log }_{10}\left(3^{0.5}\right)$ or ${\log }_{10}\left(30.5\right)$. Let's evaluate ${\log }_{10}\left(30.5\right)$ using the change of base formula and the natural logarithm values if needed, or by approximating with the given variables.

We know that:
${\log }_{10}\left(30.5\right)={\log }_{10}(3·10+0.5)$

Let's use the relation between natural log ($\ln$) and common log (${\log }_{10}$):
${\log }_{10}x=\ln \left(x\right)·{\log }_{10}e$
where ${\log }_{10}e=0.4343$.

Alternatively, we can express ${\log }_{10}30.5$ as:
${\log }_{10}30.5={\log }_{10}(10·3.05)={\log }_{10}10+{\log }_{10}3.05=1+{\log }_{10}3.05$

We can approximate ${\log }_{10}3.05$ using the first-order Taylor expansion (or linear approximation) around $x=3$:
$f\left(x\right)={\log }_{10}x$
The derivative of $f\left(x\right)$ is:
$f^{\prime }\left(x\right)=\frac{{\log }_{10}e}{x}$

Using the linear approximation formula $f(x+\Delta x)\approx f\left(x\right)+f^{\prime }\left(x\right)·\Delta x$ at $x=3$ and $\Delta x=0.05$:
${\log }_{10}\left(3.05\right)\approx {\log }_{10}3+\frac{{\log }_{10}e}{3}·0.05$

Substitute the given values ${\log }_{10}3=0.4771$ and ${\log }_{10}e=0.4343$ into the approximation:
${\log }_{10}\left(3.05\right)\approx 0.4771+\frac{0.4343}{3}·0.05$

Let's calculate the second term:
$\frac{0.4343}{3}\approx 0.14477$
$0.14477·0.05\approx 0.007238$

Now, add this to ${\log }_{10}3$:
${\log }_{10}\left(3.05\right)\approx 0.4771+0.007238=0.484338$

Finally, we find ${\log }_{10}30.5$:
${\log }_{10}30.5=1+{\log }_{10}3.05\approx 1+0.4843=1.4843$

Rounding this result to three decimal places yields the exact value of 1.484.