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 x.
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 x, since [tex]2^x>0[/tex]. [tex]log_4(2^x)=x.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 x, 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 x>0, so the only thing left here is x>0.
Since x>0, it also follows that 7x+3>0 and |7x+3|=7x+3
The equation becomes
lg8x=lg(7x+3)
8x=7x+3
x=3

Solution:
The equation is defined for x>0.
[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.\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 is equivalent to x<3 and therefore [s]28-3^x > 28-3³=28-27=1>0[/s].
The equation becomes
[tex]log_3(28-3^x)=3-x[/tex]
[tex]28-3^x=3^{3-x}[/tex]
[tex]28-3^x=\frac{27}{3^x}[/tex]
[tex]28.3^x-3^{2x}=27[/tex]
[tex]3^{2x}-28.3^x+27=0[/tex]
[tex]3^{2x}-27.3^x-3^x+27=0[/tex]
[tex]3^x(3^x-27)-(3^x-27)=0[/tex]
[tex](3^x-1)(3^x-27)=0[/tex]. Since x<3, [s]x=3 (3^x=27)[/s] is not a root of the equation, therefore the only root is gotten from [s]3^x=1[/s], or x=0.