Logarithms - Need Help!!! (1 Viewer)

skillstriker · Feb 16, 2012

Find the relationship between x and y that does not involve logarithms:

log₁₀(1+y) - log₁₀(1-y) = x

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When I try to do this question i get `y = 10^x - 10^xy - 1 BUT the answer is y = (10^x - 1) / (10^x+ 1)

AAEldar · Feb 16, 2012

$log_{10}(1+y)-log_{10}(1-y)=x$

$10^{log_{10}(1+y)-log_{10}(1-y)}=10^x$

$\frac{10^{log_{10}(1+y)}}{10^{log_{10}(1-y)}}=10^x$

$\frac{1+y}{1-y}=10^x$

$1+y=10^x(1-y)$

$1+y=10^x-y10^x$

$y+y10^x=10^x-1$

$y(1+10^x)=10^x-1$

$y=\frac{10^x-1}{10^x+1}$

Your issue (I think) was that forgetting that a minus in the index can be written as a fraction.

Jason Xie · Feb 16, 2012

log10(1+y) - log10(1-y) = x

log10[(1+Y)/(1-y)] = X

10^X = [(1+Y)/(1-y)]

skillstriker · Feb 16, 2012

Thanks!

skillstriker · Feb 17, 2012

need help with another question -
log_a2 = log_b16. Show that b = a⁴

Full working please!!
Thanks.

pokka · Feb 17, 2012

Logarithms - Need Help!!! (1 Viewer)

skillstriker

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