how do i solve these logs?

log1/2 (3x+1)^1/3= -2

3^(x^3)= 9^x

For problem 2:

3^(x^3)=9^3
take log of both sides
(x^3)log(3)=(x)log(9)
divide both sides by x
(x^2)log(3)=log(9)
9=3^2
so
(x^2)log(3)=2log(3)
divide both sides by log(3)
x^2=2
x=sqrt(2)

The first line above (where the problems was restated) should have been:

3^(x^3)=9^x
The rest of the lines are correct.

To solve these logarithmic equations, we need to understand the properties of logarithms and use algebraic techniques. Let's solve each equation step by step:

1. log(base 1/2) (3x+1)^(1/3) = -2:

To eliminate the logarithm, we can rewrite the equation in exponential form. According to the definition of logarithms, we have:

(1/2)^(-2) = (3x+1)^(1/3)

Raising both sides to the -2 power, we get:

1 / (1/2)^2 = (3x+1)^(1/3)

Simplifying:

1 / (1/4) = (3x+1)^(1/3)

4 = (3x+1)^(1/3)

To remove the cube root, we can raise both sides to the power of 3:

4^3 = (3x+1)^(1/3)^3

64 = 3x + 1

Now we can solve for x:

3x + 1 = 64

3x = 63

x = 63/3

x = 21

Therefore, the solution to the equation log(base 1/2) (3x+1)^(1/3) = -2 is x = 21.

2. 3^(x^3) = 9^x:

Since both sides have the same base (3), we can take the logarithm (log base 3) of both sides:

log(base 3) (3^(x^3)) = log(base 3) (9^x)

Using the exponent rule for logarithms, we can bring down the exponents:

x^3 * log(base 3) (3) = x * log(base 3) (9)

Log(base 3) (3) is equal to 1, and log(base 3) (9) can be simplified to 2:

x^3 = 2x

To solve this equation, we can bring all terms to one side:

x^3 - 2x = 0

Factoring out x, we get:

x * (x^2 - 2) = 0

Setting each factor equal to zero:

x = 0 or x^2 - 2 = 0

For x^2 - 2 = 0, we can solve for x:

x^2 = 2

x = ±√2

So the solutions to the equation 3^(x^3) = 9^x are x = 0 and x = ±√2.

Remember to always check your solutions by substituting them back into the original equation to ensure they are valid.