Four logarithmic equations

Statement

Solve the following equations:

• (a) $\dfrac{3}{4}-\dfrac{1}{2}\log_{1/3}(x)=-\dfrac{3}{2}$
• (b) $3-\log_4\!\left(\dfrac{x}{3}\right)=-2+3\log_4(x^2)$
• (c) $1-2\log_2(4x)+\dfrac{2}{3}\log_4\!\left(\dfrac{x^2}{2}\right)-\log_{1/3}(x)=3$
• (d) $2-\dfrac{3}{2}\log_2\!\left(x^2+3x+2\right)-3=3$

Solution

(a) Existence: $x>0$. $-\tfrac12\log_{1/3}x=-\tfrac32-\tfrac34=-\tfrac94\Rightarrow \log_{1/3}x=\tfrac92$, hence $x=\left(\tfrac13\right)^{9/2}=3^{-9/2}\approx 0.00713$.

(b) Existence: $x>0$. With $L=\log_4 x$: $3-(L-\log_4 3)=-2+6L\Rightarrow 5+\log_4 3=7L\Rightarrow L=\dfrac{5+\log_4 3}{7}\approx 0.8275$, hence $x=4^{L}\approx 3.150$.

(c) Existence: $x>0$. With $L=\log_2 x$: $\log_2(4x)=2+L$, $\log_4\!\tfrac{x^2}{2}=L-\tfrac12$, $\log_{1/3}x=-\dfrac{L}{\log_2 3}$. Substituting: $-\frac{10}{3}+\left(-2+\frac23+\frac{1}{\log_2 3}\right)L=3\ \Longrightarrow\ -0.7024\,L=\frac{19}{3},$ hence $L\approx -9.017$ and $x=2^{L}\approx 0.00193$ (a deliberately “messy” value).

(d) $-\tfrac32\log_2(x^2+3x+2)=4\Rightarrow \log_2(x^2+3x+2)=-\tfrac83$, hence $x^2+3x+2=2^{-8/3}\approx 0.1575$, i.e. $x^2+3x+1.843=0\Rightarrow x\approx -0.862\ \vee\ x\approx -2.138$. Existence: $(x+1)(x+2)>0\iff x<-2\ \vee\ x>-1$: both acceptable.

$\boxed{\text{(a) }3^{-9/2}\!\approx 0.00713;\ \text{(b) }\approx 3.150;\ \text{(c) }\approx 0.00193;\ \text{(d) }\approx -0.862\ \vee\ -2.138}$