Logarithmic Equations: Difficult Problems with Solutions

Solution:
The expression inside the brackets is [tex](3^x)^2-3^x+1=(3^x-1)^2+3^x>0[/tex], so the equation is defined for any [tex]x[/tex].
We take the base 3 antilogarithm of both sides: [tex]3^{2x}-3^x+1=3^x[/tex]
[tex]3^{2x}-2.3^x+1=0[/tex]
[tex](3^x-1)^2=0[/tex]
[tex]3^x=1[/tex]
[tex]x=0[/tex]

Solution:
We take the antilogarithm of both sides:
[tex]3x+1=16^2[/tex]
[tex]3x+1=256[/tex]
[tex]3x=255[/tex]
[tex]x=85[/tex]

Solution:
The equation is defined for all [tex]x[/tex], since [tex]2^x>0[/tex]. [tex]log_4(2^x)=x \times log_42=\frac{x}{2}[/tex]
[tex]\frac{x}{2}=5x+3[/tex]
[tex]x=10x+6[/tex]
[tex]x=-\frac{2}{3}[/tex]

Solution:
[tex]log_7(x-1)^3=6[/tex]. In order for the equation to be defined, [tex]x-1>0[/tex].
[tex]3log_7(x-1)=6[/tex]
[tex]log_7(x-1)=2[/tex]
[tex]x-1=7^2=49[/tex]
[tex]x=50[/tex]

Solution:
In order for the equation to have meaning, [tex]x>0[/tex]. Let us assume that [tex]x = 1[/tex]. Then [tex]log_5x+log_3x=log_51+log_31=0+0=0[/tex], which satisfies the equation and [tex]x=1[/tex] is a solution.
Now, let [tex]x \ne 1[/tex]. We know that
[tex]log_5x+log_3x=\frac{1}{log_x5}+\frac{1}{log_x3}=0[/tex]
[tex]log_x3+log_x5=0[/tex]
[tex]log_x(3.5)=0[/tex]
[tex]log_x15=0[/tex]
[tex]15=x^0=1[/tex], which is impossible, therefore [tex]x=1[/tex] is the only solution.

Solution:
The equation is defined for values of [tex]x[/tex], which satisfy:
[tex]\begin{array}{|l}8x>0\\7x+3 \ne 0\\lg|7x+3| \ne 0\end{array}[/tex]
[tex]\begin{array}{|l}x>0\\7x+3 \ne \pm 1\end{array}[/tex], but [tex]7x+3 > 3[/tex] follows from [tex]x>0[/tex], so the only thing left here is [tex]x>0[/tex].
Since [tex]x>0[/tex], it also follows that [tex]7x+3>0[/tex] and [tex]|7x+3|=7x+3[/tex]
The equation becomes
[tex]\lg 8x=\lg (7x+3)[/tex]
[tex]8x=7x+3[/tex]
[tex]x=3[/tex]

Solution:
The equation is defined for [tex]x>0[/tex].
[tex]log_5x-log_{25}x+log_{\sqrt{5}}x=log_5x-log_{5^2}x+log_{5^{\frac{1}{2}}}x=log_5x-\frac{1}{2}log_5x+2log_5x=\frac{5}{2}log_5x[/tex]
[tex]\frac{5}{2}log_5x=-5[/tex]
[tex]log_5x=-5 \times \frac{2}{5}=-2[/tex]
[tex]x=5^{-2}=\frac{1}{25}[/tex]

Solution:
The equation is defined for [tex]3-x>0[/tex] and [tex]28-3^x>0[/tex]. The first condition gives [tex]x<3[/tex], and then [tex]28-3^{x}>28-3^{3}=1>0[/tex], so the second holds too.
Since [tex]2^{\log_2(3-x)}=3-x[/tex], the equation becomes
[tex]\log_3(28-3^x)=3-x[/tex]
[tex]28-3^x=3^{3-x}=\frac{27}{3^x}[/tex]
[tex]28 \times 3^x-3^{2x}=27[/tex]
[tex]3^{2x}-28 \times 3^x+27=0[/tex]
[tex](3^x-1)(3^x-27)=0[/tex]
The root [tex]3^x=27[/tex], i.e. [tex]x=3[/tex], does not satisfy [tex]x<3[/tex]. Therefore [tex]3^x=1[/tex], so [tex]x=0[/tex].

Solution:
Domain: [tex]x\gt 0[/tex]. Taking logarithms to base [tex]10[/tex]:
[tex]\log ^2 x=2+\log x[/tex]; Substitute [tex]t=\log x[/tex]: [tex]t^2-t-2=0 \Rightarrow t=2, t=-1[/tex]
[tex]x_1=100, x_2=\dfrac{1}{10}[/tex]

Solution:
Domain: [tex]x\gt 0[/tex]. Substitute [tex]t=\log x[/tex]:
[tex](1+t)(t-1)=3t-3 \iff t^2-3t+2=0 \Rightarrow t=1, t=2[/tex]
[tex]x_1=10, x_2=100[/tex]

Solution:
Domain: [tex]3^x+x-17\gt 0[/tex].
[tex]x\log 30-x=x\left(\log 30-\log 10\right)=x\log 3=\log 3^x[/tex]
[tex]3^x+x-17=3^x \Rightarrow x=17[/tex]; [tex]3^{17}\gt 0\ \checkmark[/tex]

Solution:
Domain: [tex]x\gt 0[/tex]. Substitute [tex]t=\log_2 x[/tex]: [tex]\log_{\sqrt{2}}x=2t, \log_{2\sqrt{2}}x=\dfrac{2t}{3}, \log_4 x=\dfrac{t}{2}[/tex]
[tex]2t\times t\times \dfrac{2t}{3}\times \dfrac{t}{2}=\dfrac{2}{3}t^4=54 \Rightarrow t^4=81 \Rightarrow t=\pm 3[/tex]
[tex]x_1=8, x_2=\dfrac{1}{8}[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1[/tex].
[tex]\dfrac{1}{\log_x 5}=\log_5 x, x^{\log_5 2}=5^{\log_5 x\times \log_5 2}=2^{\log_5 x}[/tex]
[tex]3\times 2^{\log_5 x}+2^{\log_5 x}=4\times 2^{\log_5 x}=64 \Rightarrow 2^{\log_5 x}=16 \Rightarrow \log_5 x=4 \Rightarrow x=625[/tex]

Solution:
Domain: [tex]x+2\gt 0[/tex] ([tex]x^2+2x+4=(x+1)^2+3\gt 0[/tex]).
[tex]\left(\dfrac{1}{3}\right)^{\log_9\left(x^2+2x+4\right)}=3^{-\frac{1}{2}\log_3\left(x^2+2x+4\right)}=\dfrac{1}{\sqrt{x^2+2x+4}}[/tex]
[tex]6^{\log_{\frac{1}{6}}(x+2)}=6^{-\log_6(x+2)}=\dfrac{1}{x+2}[/tex]
[tex]\sqrt{x^2+2x+4}=x+2 \Rightarrow x^2+2x+4=x^2+4x+4 \Rightarrow x=0[/tex]

Solution:
Domain: [tex]20+x-x^2\ge 0 \iff -4\le x\le 5, 11-3x-x^2\gt 0[/tex].
[tex]20+x-x^2=0 \Rightarrow x=-4\ (11+12-16=7\gt 0\ \checkmark), x=5\ (11-15-25=-29\lt 0)[/tex]
[tex]11-3x-x^2=1 \iff x^2+3x-10=0 \Rightarrow x=2\ (20+2-4=18\ge 0\ \checkmark), x=-5\ (-5\lt -4)[/tex]
[tex](-4)\times 2=-8[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1[/tex]. Substitute [tex]t=\log_3 x\ne 0[/tex]: [tex]\log_x\left(9x^2\right)=\log_x 9+2=\dfrac{2}{t}+2[/tex]
[tex]\left(\dfrac{2}{t}+2\right)t^2=4 \iff 2t^2+2t-4=0 \iff t^2+t-2=0 \Rightarrow t=1, t=-2[/tex]
[tex]x_1=3, x_2=\dfrac{1}{9}[/tex]

Solution:
Domain: [tex]x+1\gt 0, x+1\ne 1, 2x^3+3x^2-3x+1\gt 0[/tex].
[tex]2x^3+3x^2-3x+1=(x+1)^3=x^3+3x^2+3x+1 \iff x^3-6x=0 \Rightarrow x\in\left\{0, \pm\sqrt{6}\right\}[/tex]
[tex]x=0:\ x+1=1[/tex] — is rejected.
[tex]x=-\sqrt{6}\lt -1[/tex] — is rejected.
[tex]x=\sqrt{6}:\ 2x^3+3x^2-3x+1=\left(\sqrt{6}+1\right)^3\gt 0\ \checkmark[/tex]

Solution:
Domain: [tex]x\gt -7, x\ne 5[/tex].
[tex]2\log_9(x+7)=\log_3(x+7), \log_3(x-5)^2=2\log_3|x-5|[/tex]
[tex]\log_3\dfrac{2(x+7)}{(x-5)^2}=0 \iff 2(x+7)=(x-5)^2 \iff x^2-12x+11=0[/tex]
[tex]x_1=1, x_2=11[/tex]
Both values satisfy the domain.

Solution:
Domain: [tex]x\gt 0, y\gt 0[/tex]. Taking logarithms to base [tex]10[/tex]:
[tex]\log x+\log y=\log 40=1+\log 4, \log x\times \log y=\log 4[/tex]
[tex]\log x, \log y[/tex] are the roots of [tex]z^2-(1+\log 4)z+\log 4=0 \iff (z-1)(z-\log 4)=0[/tex]
Answer: [tex](x, y)\in\{(10, 4), (4, 10)\}[/tex]

Solution:
Domain: [tex]x\gt 0, y\gt 0, x\ne 1, y\ne 1[/tex].
Substitute [tex]t=\log_x y[/tex]: [tex]t+\dfrac{1}{t}=\dfrac{5}{2} \Rightarrow t=2, t=\dfrac{1}{2}[/tex]
[tex]y=x^2:\ x^2-x^4=-2 \iff x^4-x^2-2=0 \Rightarrow x^2=2 \Rightarrow x=\sqrt{2}, y=2[/tex]
[tex]x=y^2:\ y^4-y^2+2=0[/tex] — no real roots [tex](D=-7\lt 0)[/tex].

Solution:
Domain: [tex]x\gt 0, x\ne 1, \log_x\sqrt{3x}\ge 0[/tex].
Substitute [tex]t=\log_3 x\ne 0[/tex]: [tex]\log_x\sqrt{3x}=\dfrac{1}{2}\left(\log_x 3+1\right)=\dfrac{1+t}{2t}[/tex]
The square root is non-negative and the product equals [tex]-1[/tex], so [tex]t\lt 0[/tex].
Squaring: [tex]\dfrac{1+t}{2t}\times t^2=1 \iff t^2+t-2=0 \Rightarrow t=1, t=-2[/tex]
[tex]t=-2 \Rightarrow x=\dfrac{1}{9}[/tex]
Check: [tex]\dfrac{1+t}{2t}=\dfrac{1}{4}, \sqrt{\dfrac{1}{4}}\times (-2)=-1\ \checkmark[/tex]