# Math Help - Simple indices question..

1. ## 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.

2. 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
$4^{2x} \times 8^{x-1} = 32$

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

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

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

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

Thus
7x-3 = 5

x= 8/7

3. 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.
$4^{2x}\cdot 8^{x-1} = 32$

$
2^{4x}\cdot2^{3x-3}=2^5
$