Logarithmic equations (I)

Statement

Solve the following equations after stating the existence conditions:

• (a) $\log_2(x)=3$
• (b) $2\log_3(x+1)=-1$
• (c) $\dfrac{1}{2}\log_4(x)-3=4\log_4(x)$
• (d) $3-2\log(3-x)=1$
• (e) $\dfrac{1}{3}\log_{1/2}(x)-\dfrac{5}{2}\log_2(x)=1$

Solution

(a) EC $x>0$. $x=2^3=8$.

(b) EC $x>-1$. $\log_3(x+1)=-\tfrac12 \Rightarrow x+1=3^{-1/2}=\tfrac{1}{\sqrt3}$, so $x=\tfrac{1}{\sqrt3}-1\approx-0{.}423$.

(c) EC $x>0$. $\left(\tfrac12-4\right)\log_4 x=3 \Rightarrow -\tfrac72\log_4 x=3 \Rightarrow \log_4 x=-\tfrac67$, so $x=4^{-6/7}=2^{-12/7}\approx0{.}305$.

(d) EC $x<3$. $2\log(3-x)=2 \Rightarrow \log(3-x)=1 \Rightarrow 3-x=10 \Rightarrow x=-7$.

(e) EC $x>0$. Since $\log_{1/2}x=-\log_2 x$: $\left(-\tfrac13-\tfrac52\right)\log_2 x=1 \Rightarrow -\tfrac{17}{6}\log_2 x=1 \Rightarrow \log_2 x=-\tfrac{6}{17}$, so $x=2^{-6/17}\approx0{.}783$.

$\boxed{x_a=8;\ x_b=\tfrac{1}{\sqrt3}-1;\ x_c=4^{-6/7};\ x_d=-7;\ x_e=2^{-6/17}}$