Solve the inequality $|y-5|<1$. Hence solve the inequality $|3^{x}-5|<1$, giving 3 significant figures in your answer.

1. Rewrite the modulus inequality as a double inequality.

$-1 < y - 5 < 1$

2. Solve the double inequality for $y$.

$5 - 1 < y < 5 + 1$
$4 < y < 6$

1. Solve the inequality from part (i) and apply the substitution $y = 3^x$.

First, solve part (i) concisely:
$|y-5| < 1$
$-1 < y - 5 < 1$
$4 < y < 6$
Since this part specifies "Hence", substitute $y = 3^x$ into the result from part (i):
$4 < 3^x < 6$

2. Take logarithms to bring down the exponent $x$.

Take the natural logarithm ($\ln$) or common logarithm ($\lg$) on all parts of the inequality:
$\ln 4 < \ln(3^x) < \ln 6$
Using the power rule of logarithms:
$\ln 4 < x \ln 3 < \ln 6$

3. Solve for $x$ and round to 3 significant figures.

Divide throughout by $\ln 3$ (which is positive, so the inequality signs remain unchanged):
$\frac{\ln 4}{\ln 3} < x < \frac{\ln 6}{\ln 3}$
$1.2618... < x < 1.6309...$
Rounding both boundaries to 3 significant figures, we get:
$1.26 < x < 1.63$