# Simple indices question..

• Mar 18th 2009, 01:17 AM
struck
Simple indices question..
Solve the equation 4^2x x 8^(x-1) = 32

Usually I solve these using logs and antilogs but here the question specifies that I cannot use that way as it's too basic.

I get the answer, x = 8/7 using logs/antilogs but I am not sure how to do it using a just the indices and surds rules.
• Mar 18th 2009, 01:32 AM
Quote:

Originally Posted by struck
Solve the equation 4^2x x 8^(x-1) = 32

Usually I solve these using logs and antilogs but here the question specifies that I cannot use that way as it's too basic.

I get the answer, x = 8/7 using logs/antilogs but I am not sure how to do it using a just the indices and surds rules.

4^2x x 8^(x-1) = 32

I think you mean
$\displaystyle 4^{2x} \times 8^{x-1} = 32$

$\displaystyle (2^2)^{2x} \times (2^3)^{x-1} = 2^5$

$\displaystyle 2^{4x} \times 2^{3x-3}= 2^5$

$\displaystyle 2^{4x+3x-3} = 2^5$

$\displaystyle 2^{7x-3} = 2^ 5$

Thus
7x-3 = 5

x= 8/7
• Mar 18th 2009, 01:35 AM
$\displaystyle 4^{2x}\cdot 8^{x-1} = 32$
$\displaystyle 2^{4x}\cdot2^{3x-3}=2^5$