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by **lora** » Sat Apr 13, 2013 4:24 am

Mathematics – Algebra, Exponential equations

How can we solve the following exponential equation?

$4^{6y-1}=(\frac{1}{2})^{3-5y}$

Thanks

by **miranda** » Sat Apr 13, 2013 4:35 am

Mathematics – Algebra, Exponential equations

$4^{6y-1}=(\frac{1}{2})^{3-5y}$

$(2^2)^{6y-1}=(2^{-1})^{3-5y}$

$2^{2(6y-1)}= 2^{-(3-5y)}$

$2^{12y-2)}= 2^{-3+5y}$

$12y-2= -3+5y$

$7y=1$

$y=\frac{1}{7}$

Hope these help!!

by **lora** » Sat Apr 13, 2013 4:35 am

Thanks miranda!

by **Haitham** » Wed Jan 29, 2014 9:49 am

You wrong, the answer should be -1/7

by **miranda** » Wed Jan 29, 2014 8:28 pm

Haitham wrote:You wrong, the answer should be -1/7

Yes, of course I made a mistake.
Here is the right solution:

$12y-2= -3+5y$

$7y= -1$

$y=-\frac{1}{7}$

Thanks Haitham for your comment!

by **Haitham** » Thu Apr 10, 2014 8:26 pm

Your welcome

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