Answer:
599/2
Step-by-step explanation:
[tex]\log(2x+1)-\log(6)=2[/tex]
I'm going to use quotient rule which says: [tex]\log(\frac{q}{p})=\log(q)-\log(p)[/tex]:
[tex]\log(\frac{2x+1}{6})=2[/tex]
Now we are going write this in equivalent expontial form. That is, [tex]\log_b(a)=y[/tex] implies [tex]b^y=a[/tex].
Writing in exponetial form:
[tex]10^2=\frac{2x+1}{6}[/tex]
[tex]100=\frac{2x+1}{6}[/tex]
Multiply both sides by 6:
[tex]600=2x+1[/tex]
Subtract 1 on both sides:
[tex]599=2x[/tex]
Divide both sides by 2:
[tex]\frac{599}{2}=x[/tex]
When it comes to logarithms, you should check your solution(s).
[tex]\log(2 \cdot \frac{599}{2}+1)-\log(6)=2[/tex]
[tex]\log(599+1)-\log(6)=2[/tex]
[tex]\log(600)-\log(6)=2[/tex]
[tex]\log(\frac{600}{6})=2[/tex]
[tex]\log(100)=2[/tex]
[tex]2=2[/tex]
The solution checks out.
Answer:
x=299.5
Step-by-step explanation:
I will convert log(6) to its decimal form
[tex]log(2x+1)-log(6)=2\\\\log(2x+1)-0.778151=2[/tex]
Step 1: Add 0.778151 to both sides.
[tex]log(2x+1)-0.778151+0.778151=2+0.778151\\\\log(2x+1)=2.778151[/tex]
Step 2: Solve Logarithm.
[tex]log(2x+1)=2.778151\\\\10^{log(2x+1)}=10^{102.778151}[/tex] Take the exponent of both sides.
[tex]2x+1=102.778151\\\\2x+1=599.999654\\\\2x+1-1=599.999654-1[/tex]
Subtract 1 from both sides.
[tex]2x=598.999654\\\\\\\frac{2x}{2}=\frac{598.999654}{2}[/tex]
Divide both sides by 2.
[tex]x=299.5[/tex]
