Solve the following equations. Round to 3 decimal places.

Question

a) 2^x+1 = 3^x-1

b) log(2x+1) + log(x-1)= log9

Answer 1

a) To get x out of the exponent, we need to take a log or natural log of both sides (your answer will turn out the same either way). Using properties of exponents (specifically x^nm = xⁿ * x^m) and properties of logs (specifically log(mn) = log(m) + log(n) and log(m/n) = log(m)-log(n)), we can isolate and solve for x.

 

b) Using properties of logs (specifically log(mn) = log(m) + log(n)), we can combine the two left side terms into one term. Then, take 10 to the power of both sides. And then finally, we can solve for x.

Answer 2

b)  I would like to show another approach we can use to solve for x than just getting the logs to one side of equation and using the base number 10.

 

log(2x+1) + log(x-1)= log9

 

 

When we have sum of logs, we can rewrite as a product as Andrew mentioned.

 

log[(2x + 1)(x - 1)] = log9

 

 

Notice that we have log on both sides of equation.  We can set the argument of log on the left side equal to 9 so that the left side will be log9.  This makes log9 = log9.

 

(2x + 1)(x - 1) = 9

 

2x² - x - 1 = 9

 

 

Subtract 9 on both sides of equation.

 

2x² - x - 10 = 0

 

 

Factor the left side of equation.

 

(2x - 5)(x + 2) = 0

 

 

Set the factors equal to zero.

 

2x - 5 = 0        and          x + 2 = 0

 

x = 5/2            and          x = -2

 

 

Substitute these values of x to validate them.  Since we cannot have a negative value in the argument of a log, we accept the positive value of x.

 

x = 5/2    

 

or  

 

x = 2.5