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.

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- Oct 3rd 2009, 12:46 AMyeokyLogarithm question
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.

- Oct 3rd 2009, 03:11 AMpacman
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 - Oct 3rd 2009, 03:14 AMMatt Westwood
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 $\displaystyle a = b^{\log_b a}$. - Oct 3rd 2009, 09:44 PMyeoky
- Oct 4th 2009, 02:01 AMpacman
y = 8(27^log_6 X)+27(8^log_6 X) - 6^(log_6 x^3),

y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 6^3(log_6 x)

y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 216^(log_6 x)'

y = ? - Oct 4th 2009, 02:18 AMyeoky