Need help understanding how to solve logarithmic equations for my test tommorrow.

Question

1) 2x =4 ^3√8

 

2) 5^x=76

 

3) log₃1/27=x

 

4)2^3x-1=30

 

thanks it's much appreciated!

Answer 1

For question 3, we can also rewrite as a difference of logs, since we are dividing the arguments.

 

x = log₃(1/27)

 

x = log₃(1) - log₃(27)

 

3 raised to what power results in 1?  That would be 0.

3 raised to what power results in 27?  That would be 3.

 

Therefore,

 

x = 0 - 3

x = -3

Answer 2

I do not know what base you want to work in, but here are valid expressions:

 

(1)

 

2x = 4 * 8^(1/3) = 4 * (2^3)^(1/3) = 4*2 = 8, so x = 8/2 = 4

 

(2)

 

5^x = 76

 

x*log(5) = log(76)

 

x = log(76) / log(5), true for any base

 

(3)

 

x = log₃(1/27) = log₃[(3^3)^(-1)] = log₃[(3)^(-3)] = -3 log₃(3) = -3 (1) = -3

 

(4)  

 

2^(3x-1) = 30

 

(3x-1) log(2) = log(30)

 

x = (1/3){[log(30) / log(2)]  + 1}, true for any base