Solve.
log(x^3)-log(2)=log(2x^2)
3logx-log2=2logx*log2
3logx-2logx log2=log2
logx(3-2log2)=log2
logx= log2/(3-2log2)
look up log2, and solve.
I don't think that is right
answer is x=4
not sure how
how about this:
log(x^3)-log(2)=log(2x^2)
log [x^3/2] = log(2x^2)
x^3/2 = 2x^2
x^3 = 4x^2
x^3 - 4x^2 = 0
x^2(x-4) = 0
x = 0, not possible
or
x = 4
To solve the equation log(x^3) - log(2) = log(2x^2), we will use the properties of logarithms.
Step 1: Combine the logarithms on the left side using the quotient rule of logarithms. According to the quotient rule, log(a) - log(b) = log(a/b).
log(x^3) - log(2) = log(x^3 / 2) = log(2x^2)
Step 2: We can now set the bases equal to each other, as the logarithmic functions have the same base. In this case, the base is 10 (common logarithm).
x^3 / 2 = 2x^2
Step 3: Eliminate the fraction by multiplying both sides of the equation by 2.
2(x^3 / 2) = 2(2x^2)
x^3 = 4x^2
Step 4: Rearrange the equation to bring all terms to one side, in this case, the right side.
x^3 - 4x^2 = 0
Step 5: Factor out the common term "x^2" from both terms.
x^2(x - 4) = 0
Step 6: Set each factor equal to zero and solve for x.
x^2 = 0 or x - 4 = 0
If x^2 = 0, then x = 0.
If x - 4 = 0, then x = 4.
Therefore, the solutions to the equation log(x^3) - log(2) = log(2x^2) are x = 0 and x = 4.