That would be great if you could show all the steps. Thanks! :D
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10^(2x) = (10^x)²
Let t = 10^x so t > 0 because 10^x is always greater than 0
t² + 3t - 10 = 0
(t + 5)(t - 2) = 0
t = -5 is not the answer
t = 2
10^x = 2
take log
x = log2
x is about 0.301
Let t = 10^x so t > 0 because 10^x is always greater than 0
t² + 3t - 10 = 0
(t + 5)(t - 2) = 0
t = -5 is not the answer
t = 2
10^x = 2
take log
x = log2
x is about 0.301
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10^2x can be rewritten
10²*10^x
let 10^x=V
the equation is now
10²V+3V-10=0
solve for V, replace V=10^x, solve for x
10²*10^x
let 10^x=V
the equation is now
10²V+3V-10=0
solve for V, replace V=10^x, solve for x
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This is reducible to a quadratic.
Let u = 10^x
u^2 + 3u - 10 = 0
(u-2)(u+5) = 0
u = 2 or -5
10^x = 2 or 10^x = -5
Sorry, i don't know further from here since I havent done logarithms.
Let u = 10^x
u^2 + 3u - 10 = 0
(u-2)(u+5) = 0
u = 2 or -5
10^x = 2 or 10^x = -5
Sorry, i don't know further from here since I havent done logarithms.