# Logarithm question

• Oct 3rd 2009, 12:46 AM
yeoky
Logarithm 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 AM
pacman
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 AM
Matt Westwood
Quote:

Originally Posted by yeoky
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.

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 PM
yeoky
Quote:

Originally Posted by yeoky
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.

Thanks and appreciated and I fully understand the answer.

By the way, is there way to prove that the max is 216 besides using the graph or guess and check method.
• Oct 4th 2009, 02:01 AM
pacman
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 AM
yeoky
Quote:

Originally Posted by pacman
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 = ?

Thanks again. I understand this. But it is like gussing x=1, then y=216, but we will not know unless the graph is plotted or a few values of x are tested. Is there any way to prove or show?