Five logarithmic equations

Statement

Solve the following equations:

• (a) $\log_{1/2}(x)=-\dfrac{1}{2}$
• (b) $\log_4(x^2)=-2$
• (c) $1-2\log_{1/2}(x)=4+3\log_{1/2}(x)$
• (d) $3\log_2\left(\dfrac{4}{x}\right)+2\log_3(x^2)-\log_{1/2}(4x)=1$
• (e) $2\log\left(2x^2-18\right)+3=0$

Solution

(a) Switch to exponential form: $x=\left(\tfrac12\right)^{-1/2}=2^{1/2}=\sqrt{2}\approx 1.414$ (acceptable, $x>0$).

(b) Existence: $x\ne 0$. From $x^2=4^{-2}=\tfrac{1}{16}$ we get $x=\pm\tfrac14$. Both acceptable: $x=\pm 0.25$.

(c) Collect the logarithms: $1-4=3\log_{1/2}x+2\log_{1/2}x\Rightarrow -3=5\log_{1/2}x\Rightarrow \log_{1/2}x=-\tfrac35$. Hence $x=\left(\tfrac12\right)^{-3/5}=2^{3/5}\approx 1.516$.

(d) Existence: $x>0$. Put $L=\log_2 x$ and rewrite everything in base $2$: $3(2-L)+\frac{4L}{\log_2 3}+(2+L)=1\ \Longrightarrow\ 8+\left(\frac{4}{\log_2 3}-2\right)L=1.$ Hence $L=\dfrac{-7}{\frac{4}{\log_2 3}-2}\approx -13.37$ and so $x=2^{L}\approx 9.4\cdot 10^{-4}$ (an acceptable but deliberately “messy” value: the bases $2$, $3$ and $\tfrac12$ do not combine cleanly).

(e) The logarithm is base $10$. $\log(2x^2-18)=-\tfrac32\Rightarrow 2x^2-18=10^{-3/2}\approx 0.0316$, so $x^2\approx 9.016$ and $x\approx\pm 3.003$. Existence: $2x^2-18>0\iff |x|>3$: both acceptable.

$\boxed{\text{(a) }\sqrt2;\ \text{(b) }\pm\tfrac14;\ \text{(c) }2^{3/5};\ \text{(d) }x=2^{-7/\left(\frac{4}{\log_2 3}-2\right)}\!\approx 9.4\cdot 10^{-4};\ \text{(e) }\pm 3.003}$