Rewrite the equation [tex]\log (\alpha \beta)=5 x^2+8 y^2[/tex] in the form [tex]10^x=y[/tex] using exponents instead of logarithms.

[tex]\alpha \beta = 10^{5 x^2 + 8 y^2}[/tex]



Answer :

Certainly! Let's transform the given equation [tex]\(\log (\alpha \beta) = 5x^2 + 8y^2\)[/tex] into an exponential form.

1. Understand the Logarithmic Equation:
The given equation is:
[tex]\[ \log (\alpha \beta) = 5x^2 + 8y^2 \][/tex]
This can be interpreted as a logarithm with base 10 if no base is specified.

2. Rewrite Using the Definition of Logarithms:
Recall that the equation [tex]\(\log_b(a) = c\)[/tex] is equivalent to [tex]\(a = b^c\)[/tex]. Here the base [tex]\(b\)[/tex] is assumed to be 10.

3. Apply the Exponential Definition:
According to the definition, [tex]\(\log (\alpha \beta) = 5x^2 + 8y^2\)[/tex] implies that:
[tex]\[ \alpha \beta = 10^{5x^2 + 8y^2} \][/tex]

Thus, the equation [tex]\(\log (\alpha \beta) = 5x^2 + 8y^2\)[/tex] can be rewritten in exponential form as:
[tex]\[ \alpha \beta = 10^{5x^2 + 8y^2} \][/tex]

So, the solution in the required form is:
[tex]\[ \alpha \beta = 10^{5 \cdot x^2 + 8 \cdot y^2} \][/tex]

Thus, we have:
[tex]\[ \alpha \beta = 10^{5x^2 + 8y^2} \][/tex]