Logarithmic Equations: Problems with Solutions

Solution:
The equation is defined for [tex]x+2>0[/tex].
We raise 2 to the power of each side of the equation. The resulting equation is
[tex]2^{\log_2{x+2}}=2^3[/tex]
[tex]x+2=8[/tex]
[tex]x=6.[/tex]

Solution:
We take the base 9 antilogarithm:
[tex]3^x=9^{15}[/tex]
[tex]3^x=3^{30}[/tex]
[tex]x=30[/tex]

Solution:
We take the base 5 antilogarithm of both sides:
[tex]5^{log_5x}=5^3[/tex]
[tex]x=125[/tex]

Solution:
The logarithm function is defined for [tex]x > 0, x \ne 1[/tex].
[tex]36=x^2[/tex]
[tex]x = \pm 6[/tex], but [tex]x>0[/tex], therefore [tex]x=6[/tex] is the only solution.

Solution:
We take the base 9 antilogarithm:
[tex]x=9^{\frac{1}{2}}=\sqrt{9}=3[/tex]

Solution:
The equation is defined for [tex]x^2>0[/tex], equivalent to [tex]x \ne 0[/tex]. We take the antilogarithm:
[tex]x^2=5^6[/tex]
[tex]x^2-15625=0[/tex], which is a quadratic equation with non-zero roots (they are both roots to the logarithmic equation), therefore by Vieta\'s formulas the product is [tex]-15625[/tex].

Solution:
[tex]x^3=5^{12}[/tex]
[tex]x=5^4=625[/tex]

Solution:
[tex]x > 0, x \ne 1[/tex]
[tex]x^{\frac{1}{4}}=\sqrt{3}[/tex]
[tex]x=\sqrt{3}^4=9[/tex]

Solution:
In order for the equation to be defined, the following must be true: [tex]x^{13}>0[/tex], [tex]x>0[/tex].
[tex]13log_{13}x=26[/tex]
[tex]log_{13}x=2[/tex]
[tex]x=13^2=169[/tex]

Solution:
[tex]log_525=log_5(5^2)=2log_55=2[/tex]
[tex]2=2x => x=1[/tex]

Solution:
Domain: [tex]x\gt 0[/tex].
[tex]0. 25^{-1}=4 \Rightarrow \log_{0. 25}4=-1 \Rightarrow -\log_{0. 25}4=1=\log_5 5[/tex]
[tex]\dfrac{1}{2}\log_5 49=\log_5 7[/tex]
[tex]\log_5 x=\log_5\dfrac{5\times 350}{7}=\log_5 250 \Rightarrow x=250[/tex]

Solution:
[tex]\log_2\left(\log_{\sqrt{5}}x\right)=4^{\frac{1}{2}}=2 \Rightarrow \log_{\sqrt{5}}x=2^2=4 \Rightarrow x=\left(\sqrt{5}\right)^4=25[/tex]
Check: [tex]\log_{\sqrt{5}}25=4\gt 0, \log_2 4=2\gt 0, \log_4 2=\dfrac{1}{2}\ \checkmark[/tex]

Solution:
Domain: [tex]7x+2\gt 0, 5x-2\gt 0 \Rightarrow x\gt \dfrac{2}{5}[/tex].
[tex]\log_5(5x-2)\times \log_2 5=\log_2(5x-2)[/tex]
[tex]\log_2\dfrac{7x+2}{5x-2}=1 \Rightarrow 7x+2=2(5x-2) \Rightarrow x=2[/tex]
Check: [tex]\log_2 16-\log_2 8=4-3=1\ \checkmark[/tex]

Solution:
Domain: [tex]x^2-4x-5\gt 0, 7-3x\gt 0 \Rightarrow x\lt -1[/tex].
[tex]2\log_9(7-3x)=\log_3(7-3x)[/tex]
[tex]x^2-4x-5=7-3x \iff x^2-x-12=0 \Rightarrow x=4, x=-3[/tex]
[tex]x=4[/tex] is not in the domain.
Answer: [tex]x=-3[/tex]

Solution:
Domain: [tex]x\gt -1[/tex].
[tex]x^2-3=0 \Rightarrow x=\pm\sqrt{3}[/tex]; [tex]x=-\sqrt{3}\lt -1[/tex] is not in the domain.
[tex]\log_3(x+1)=0 \Rightarrow x+1=1 \Rightarrow x=0[/tex]
[tex]x_1=0, x_2=\sqrt{3}[/tex]

Solution:
Substitute [tex]u=x^2+4x-5\gt 0[/tex]: [tex]\dfrac{\ln u}{\ln 3}=\dfrac{\ln u}{\ln 5} \iff \ln u\left(\dfrac{1}{\ln 3}-\dfrac{1}{\ln 5}\right)=0[/tex]
[tex]\ln u=0 \Rightarrow u=1 \iff x^2+4x-6=0 \Rightarrow x=-2\pm\sqrt{10}[/tex]

Solution:
[tex]x^3-3x+5=x+5\gt 0 \iff x^3-4x=0, x\gt -5[/tex]
[tex]x(x-2)(x+2)=0 \Rightarrow x\in\{-2, 0, 2\}[/tex]

Solution:
Domain: [tex]x\gt 0, y\gt 0[/tex].
[tex]\log (xy)=\log 14 \Rightarrow xy=14[/tex]
[tex]x=y+5:\ y^2+5y-14=0 \Rightarrow y=2, y=-7[/tex]
[tex](7, 2)\ \checkmark[/tex]; [tex](-2, -7)[/tex] is not in the domain.

Solution:
Domain: [tex]x-3\gt 0, x-3\ne 1 \iff x\gt 3, x\ne 4[/tex] ([tex]x^2-4x+7=(x-2)^2+3\gt 0[/tex]).
[tex](x-3)^2=x^2-4x+7 \iff -6x+9=-4x+7 \Rightarrow x=1[/tex]
[tex]x=1[/tex] is not in the domain — no solution.

Solution:
Domain: [tex]x\gt 2[/tex].
The left side is strictly increasing and the right side is strictly decreasing, so the equation has at most one root.
[tex]x=5:\ \log_3 3=1=6-5\ \checkmark[/tex]
Answer: [tex]x=5[/tex]