SOLUTION: what is the inverse of y= log2 (8^(x-1))

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Question 941697: what is the inverse of y= log2 (8^(x-1))
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
to find the inverse of y=++log%282%2C8%5E%28x-1%29%29, first swap x and y
x=++log%282%2C8%5E%28y-1%29%29...now solve for y
x=++log%282%2C8%5E%28y-1%29%29...change to base 10
x=++log%288%5E%28y-1%29%29%2Flog%282%29
x%2Alog%282%29=++log%288%5E%28y-1%29%29
log%282%5Ex%29=++log%288%5E%28y-1%29%29...since log same, we have
2%5Ex=+8%5E%28y-1%29....we can write 8 as 2%5E3
2%5Ex=+%282%5E3%29%5E%28y-1%29
2%5Ex=+2%5E%283%28y-1%29%29.....if bases same then exponents are same too
x=+3%28y-1%29
x=+3y-3
x%2B3=+3y
y=x%2F3%2B1..=>the inverse of y=++log%282%2C8%5E%28x-1%29%29