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[/tex] => [tex]x=1[/tex]