SOLUTION: Solve each equation for x A. 27^X=3^X+1 B. e^3x-1=12 C.log(x+3)-log x=log 2 D.log5 X+ 2log5 X=3 E.ln(X+5)-ln 3=0

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve each equation for x A. 27^X=3^X+1 B. e^3x-1=12 C.log(x+3)-log x=log 2 D.log5 X+ 2log5 X=3 E.ln(X+5)-ln 3=0      Log On


   

Question 166638: Solve each equation for x
A. 27^X=3^X+1
B. e^3x-1=12
C.log(x+3)-log x=log 2
D.log5 X+ 2log5 X=3
E.ln(X+5)-ln 3=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started.


A)

27%5E%28x%29=3%5E%28x%2B1%29 Start with the given equation


%283%5E3%29%5E%28x%29=3%5E%28x%2B1%29 Rewrite 27 as 3%5E3


3%5E%283x%29=3%5E%28x%2B1%29 Multiply the exponents.


Since the bases are equal, this means that the exponents are equal


3x=x%2B1 Set the exponents equal to one another.


3x-x=1 Subtract x from both sides.


2x=1 Combine like terms on the left side.


x=%281%29%2F%282%29 Divide both sides by 2 to isolate x.


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Answer:

So the answer is x=1%2F2 which in decimal form is x=0.5.





B)

e%5E%283x-1%29=12 Start with the given equation


3x-1=ln%2812%29 Take the natural log of both sides to eliminate the base "e"


3x=ln%2812%29%2B1 Add 1 to both sides.


x=%28ln%2812%29%2B1%29%2F3 Divide both sides by 3 to isolate "x"



So the answer is x=%28ln%2812%29%2B1%29%2F3 which approximates to 1.16164