Hi can someone solve this for me.2logy = log2 + 3logx, 2 ^ y = 4 ^ x
taking antilog ... y^2 = 2 x^3
2^y = (2^2)^x ... 2^y = 2^(2x) ... y = 2 x
substituting ... 4 x^2 = 2 x^3
... solve for x , then substitute back to find y
To solve this equation, let's break down the problem step by step.
1. Start with the equation 2logy = log2 + 3logx.
2. Use the logarithmic property log(ab) = log(a) + log(b) to simplify the equation. Rewrite log2 as log(2^1) to get:
2logy = log(2^1) + 3logx
2logy = log(2) + log(x^3)
3. Apply another logarithmic property log(a^n) = nlog(a) to further simplify the equation:
2logy = log(2) + 3log(x^3)
2logy = log(2) + 3(3log(x))
4. Combine like terms:
2logy = log(2) + 9log(x)
5. Now, convert the equation with the same base logarithms (base 2 is commonly used):
2logy = log(2) + log(x^9) (using the property log(ab) = log(a) + log(b))
6. Apply the property log(a) + log(b) = log(ab) to rewrite the equation:
2logy = log(2x^9)
7. Since we have the same base logarithms, we can equate the exponents:
2y = 2x^9
8. Divide both sides of the equation by 2:
y = x^9
9. Lastly, use the given equation 2^y = 4^x to substitute x^9 for y:
2^(x^9) = 4^x
Now you have the simplified equation in terms of x.