Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.
Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.
a^logaX^3 = x^3,
Let y = 8(27^log_6 X)+27(8^log_6 X)-X^3
y = 8(27^log_6 X)+27(8^log_6 X) - 6^(log_6 x^3)
= 216 when x = 6,
THUS, global maximum is when x = 6.
i plot it first and by observation, the logarithmic function gets a maximum @ about x = 6. So i guess, max when x = 6.
ok
Logarithms are popular this time of year ...
I presume the equation you are trying to solve is:
$\displaystyle a^{\log_a (X^3)} $.
Remember how logarithms are defined. For any given base (in this case it's $\displaystyle a$), the "logarithm" function and the "exponentiation" functions are inverses of each other.
So to "undo" a logarithm, you take the exponent. To "undo" an exponent, you take its logarithm.
So $\displaystyle a^{\log_a x} = \log_a (a^x)$.
Let's do it in more detail. Suppose $\displaystyle a = b^c$. Then by definition of logarithm, $\displaystyle \log_b a = c$.
Because $\displaystyle c = \log_b a$ you can slot it straight into the equation $\displaystyle a = b^c$ to get $\displaystyle a = b^{\log_b a}$.