Exponential equations

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Exponential equations

Mathematics – Algebra, Exponential equations

How can we solve the following exponential equation?

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

Thanks
lora

Posts: 0
Joined: Wed Apr 10, 2013 7:36 pm

Re: Exponential equations

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!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

Re: Exponential equations

Thanks miranda!
lora

Posts: 0
Joined: Wed Apr 10, 2013 7:36 pm

Re: Exponential equations

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

Posts: 0
Joined: Tue Jan 28, 2014 8:15 pm

Re: Exponential equations

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

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

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

$7y= -1$

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

miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm