Well if you look at it this way 1 +1 wouldnt necessarily = 3 either it could iqual 4,5,6,7 etc. Take 1 male and 1 female and put them together and with in a short time you will end up with 3 you normally end up with at least 1 baby so the 1 and the 1 ended up being 3. 1600 9 1/3 4 2/3 56 hrs 28 hrs 14 hrs 7 hrs 3.5 hrs 2.5 hrs 1.75 hrs 1.5 hrs 1.25 hrs 1 hr 45 min 1500 8 2/3 4 1/3 52 hrs 26 hrs 13 hrs 6.5 hrs 3.25 hrs 2.25 hrs 1.5 hrs 1.25 hrs 1 hr 1 hr 45 min.
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where [latex]bne 1[/latex],
[latex]{mathrm{log}}_{b}S={mathrm{log}}_{b}Ttext{ if and only if }S=T[/latex].
For example,
[latex]text{If }{mathrm{log}}_{2}left(x - 1right)={mathrm{log}}_{2}left(8right),text{then }x - 1=8[/latex].
So, if [latex]x - 1=8[/latex], then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: [latex]{mathrm{log}}_{2}left(9 - 1right)={mathrm{log}}_{2}left(8right)=3[/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation [latex]mathrm{log}left(3x - 2right)-mathrm{log}left(2right)=mathrm{log}left(x+4right)[/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x:
[latex]begin{cases}mathrm{log}left(3x - 2right)-mathrm{log}left(2right)=mathrm{log}left(x+4right)hfill & hfill text{ }mathrm{log}left(frac{3x - 2}{2}right)=mathrm{log}left(x+4right)hfill & text{Apply the quotient rule of logarithms}.hfill text{ }frac{3x - 2}{2}=x+4hfill & text{Apply the one to one property of a logarithm}.hfill text{ }3x - 2=2x+8hfill & text{Multiply both sides of the equation by }2.hfill text{ }x=10hfill & text{Subtract 2}xtext{ and add 2}.hfill end{cases}[/latex]
To check the result, substitute x Network radar monitor your network 2 8 1. = 10 into [latex]mathrm{log}left(3x - 2right)-mathrm{log}left(2right)=mathrm{log}left(x+4right)[/latex].
[latex]begin{cases}mathrm{log}left(3left(10right)-2right)-mathrm{log}left(2right)=mathrm{log}left(left(10right)+4right)hfill & hfill text{ }mathrm{log}left(28right)-mathrm{log}left(2right)=mathrm{log}left(14right)hfill & hfill text{ }mathrm{log}left(frac{28}{2}right)=mathrm{log}left(14right)hfill & text{The solution checks}.hfill end{cases}[/latex]
A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
Direct message for instagram 4 1 1. For any algebraic expressions S and T and any positive real number b, where [latex]bne 1[/latex],
[latex]{mathrm{log}}_{b}S={mathrm{log}}_{b}Ttext{ if and only if }S=T[/latex]
Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
How To: Given an equation containing logarithms, solve it using the one-to-one property.
Example 12: Solving an Equation Using the One-to-One Property of Logarithms![]() Proper Volume 1 1 3 Divided By 2
Solve [latex]mathrm{ln}left({x}^{2}right)=mathrm{ln}left(2x+3right)[/latex].
Solution1 1/3 As A Decimal
[latex]begin{cases}text{ }mathrm{ln}left({x}^{2}right)=mathrm{ln}left(2x+3right)hfill & hfill text{ }{x}^{2}=2x+3hfill & text{Use the one-to-one property of the logarithm}.hfill text{ }{x}^{2}-2x - 3=0hfill & text{Get zero on one side before factoring}.hfill left(x - 3right)left(x+1right)=0hfill & text{Factor using FOIL}.hfill text{ }x - 3=0text{ or }x+1=0hfill & text{If a product is zero, one of the factors must be zero}.hfill text{ }x=3text{ or }x=-1hfill & text{Solve for }x.hfill end{cases}[/latex]
Analysis of the Solution
There are two solutions: x = 3 or x = –1. The solution x = –1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
Try It 12Proper Volume 1 1 3 5 6 Show Work
Solve [latex]mathrm{ln}left({x}^{2}right)=mathrm{ln}1[/latex].
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