Logarithmic equations (III)

Statement

Solve the following equations after stating the existence conditions:

• (a) $\dfrac{2}{3}\log_{1/2}(x)=-4$
• (b) $3-\dfrac{1}{3}\log_3(2-3x)=0$
• (c) $\dfrac{1}{2}\log_5(x)-3=4\log_3(x)$
• (d) $\dfrac{2}{3}\log_{1/4}(x)+\dfrac{5}{2}\log_3(x)=\dfrac{1}{2}$

Solution

(a) EC $x>0$. $\log_{1/2}x=-6 \Rightarrow x=\left(\tfrac12\right)^{-6}=2^{6}=64$.

(b) EC $x<\tfrac23$. $\tfrac13\log_3(2-3x)=3 \Rightarrow \log_3(2-3x)=9 \Rightarrow 2-3x=3^{9}=19683 \Rightarrow x=\dfrac{2-19683}{3}=-\dfrac{19681}{3}\approx-6560{.}33$.

(c) EC $x>0$. Switching to natural logarithms: $\left(\dfrac{1}{2\ln 5}-\dfrac{4}{\ln 3}\right)\ln x=3 \Rightarrow (-3{.}3303)\ln x=3 \Rightarrow \ln x\approx-0{.}9008$, so $x\approx0{.}406$.

(d) EC $x>0$. With $\log_{1/4}x=-\tfrac12\log_2 x$ and switching to natural logarithms one gets $1{.}7947\,\ln x=0{.}5 \Rightarrow \ln x\approx0{.}2786$, so $x\approx1{.}321$.

$\boxed{x_a=64;\ x_b=-\tfrac{19681}{3}\approx-6560{.}33;\ x_c\approx0{.}406;\ x_d\approx1{.}321}$