a) $$ \log _{2}(y)-2 \log _{2}(25)=2 $$ b) $$ 2,4^{x}=3,9 $$ c) $$ 5^{1-2 x}=17 $$ d) $$ 5^{2 y}=4^{1-y} $$ e) $$ 3 \cdot 1,4^{3 x}=2^{x-1} $$

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a)
$$
\log_{2}(y)-2\log_{2}(25)=2
$$
• Use the logarithm property: $$\log a - \log b = \log\left(\frac{a}{b}\right)$$ and the power rule: $$2\log_{2}(25)=\log_{2}(25^2)$$. Then,
$$
\log_{2}\left(\frac{y}{25^2}\right)=\log_{2}\left(\frac{y}{625}\right)=2.
$$
• Rewrite in exponential form:
$$
\frac{y}{625}=2^2=4.
$$
• Solve for $$y$$:
$$
y=4\cdot625=2500.
$$

b)
$$
2.4^{\,x}=3.9
$$
• Take the natural logarithm of both sides:
$$
\ln\left(2.4^{\,x}\right)=\ln(3.9).
$$
• Use the logarithm power rule:
$$
x\ln(2.4)=\ln(3.9).
$$
• Solve for $$x$$:
$$
x=\frac{\ln(3.9)}{\ln(2.4)}.
$$

c)
$$
5^{\,1-2x}=17.
$$
• Take the natural logarithm of both sides:
$$
\ln\left(5^{\,1-2x}\right)=\ln(17).
$$
• Bring the exponent in front:
$$
(1-2x)\ln5=\ln17.
$$
• Solve for $$x$$:
$$
1-2x=\frac{\ln17}{\ln5} \quad\Longrightarrow\quad 2x=1-\frac{\ln17}{\ln5},
$$
$$
x=\frac{1}{2}-\frac{\ln17}{2\ln5}.
$$

d)
$$
5^{2y}=4^{\,1-y}.
$$
• Take the natural logarithm of both sides:
$$
\ln\left(5^{2y}\right)=\ln\left(4^{\,1-y}\right).
$$
• Bring the exponents forward:
$$
2y\ln5=(1-y)\ln4.
$$
• Expand and collect the $$y$$-terms:
$$
2y\ln5= \ln4-y\ln4,
$$
$$
2y\ln5+y\ln4=\ln4,
$$
$$
y(2\ln5+\ln4)=\ln4.
$$
• Solve for $$y$$:
$$
y=\frac{\ln4}{2\ln5+\ln4}.
$$

e)
$$
3\cdot 1.4^{3x}=2^{x-1}.
$$
• Take the natural logarithm of both sides:
$$
\ln\left(3\cdot 1.4^{3x}\right)=\ln\left(2^{x-1}\right).
$$
• Use logarithm properties:
$$
\ln3+3x\ln1.4=(x-1)\ln2.
$$
• Expand the right side:
$$
\ln3+3x\ln1.4=x\ln2-\ln2.
$$
• Gather the terms involving $$x$$:
$$
3x\ln1.4-x\ln2=-\ln2-\ln3,
$$
$$
x(3\ln1.4-\ln2)=-\ln6.
$$
• Solve for $$x$$:
$$
x=\frac{-\ln6}{3\ln1.4-\ln2}.
$$
a) $$ y = 2500 $$ b) $$ x = \frac{\ln(3.9)}{\ln(2.4)} $$ c) $$ x = \frac{1}{2} - \frac{\ln(17)}{2\ln(5)} $$ d) $$ x = \frac{\ln(4)}{2\ln(5) + \ln(4)} $$ e) $$ x = \frac{-\ln(6)}{3\ln(1.4) - \ln(2)} $$

a)
b)
c)
d)
e)

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