log

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given that
4^(2x+1) X 7^(x-1) = 8^x X 14^2x
evaluate 14^x

this is what i did, but i got stuck anyways.

2^(4x+2) X 7^(x-1) = 2^3x X 14^2x
2^(x+2) = 2^2x X 7^(x+1)

sorry this might be a bit confusing, but the big letter X is 'multiply'

• log -

4^(2x+1) * 7^(x-1) = 8^x * 14^2x
4^2x * 4 * 7^x /7 = (4*2)^x * (7^x)^2
(4/7)*(4^x)^2 = 4^x * 2^x * 7^x
(4/7)* 4^x = 2^x * 7^x
(4/7)* (2*2)^x = 2^x * 7^x = (14^x)
(4/7)*2^x = 14^x = (2*7)^x
(4/7) = 7^x
x log 7 = log (4/7) = -.559616
x = -.28758
14^x = 0.46816

Check my thinking. There may be a mistake along the way.

I used base e log to get x, but the base doesn't matter in this case.

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