Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Exponential Equation 42^2x+1=1024 I don't understand how to work out the exponent 2x+1

OpenStudy (jdoe0001):

have you done logarithms yet?

OpenStudy (anonymous):

No, not yet.

OpenStudy (anonymous):

I am in Algebra 1 if that helps...

OpenStudy (jdoe0001):

hmm maybe it's just me... but I don't see it coming up without logarithms

OpenStudy (judygreeneyes):

That's the first thing I thought of also, jdoe0001, logarithms.

OpenStudy (jdoe0001):

yeap

OpenStudy (jdoe0001):

yeah... I'd think you have to use logs for it

OpenStudy (anonymous):

It is. You take the log base 42: \[42^{2x+1}=1024 \implies \log_{42}(42^{2x+1})=\log_{42}(1024) \implies 2x+1=\log_{42}(1024)\] \[\implies x=\frac{\log_{42}(1024)-1}{2}\]

OpenStudy (anonymous):

What is a log? @malevolence19 @jdoe0001 @judygreeneyes

OpenStudy (judygreeneyes):

Yikes! If they haven't covered logarithms yet, that's pretty complicated (and correct)

OpenStudy (anonymous):

|dw:1389652809977:dw| this is what i did

OpenStudy (judygreeneyes):

A logarithm answers the question "How many of one number do we multiply to get another number?" Before we had easy access to big computers, laptops, and scientific calculators, we used logarithm tables to help us handle calculations involving large exponents. For the case of base 10 (our usual numbering system) a simple example is log(100) = ? (sometimes this would be written, for clarity, as \[\log _{10}(100)\] to show that the question refers to base 10). So the question really is, what exponent is needed to get the base number (10) to equal 100. We know that 10^2 = 100 So an exponent of 2 is needed to make 10 into 100, and therefore \[\log _{10}(100) = 2\] So in your problem, to get that nasty 2x+1 out of the exponent, you can take the logarithm (base 42) of both sides of the equation. The \[\log _{42}\] of 42^(2x+1) is just the exponent 2x+1. The other side of the equation becomes \[\log _{42}\] of 1024, and then you solve for X. You can treat the log of 1024 like a number since it really is one. Using Excel, I got \[\log _{42}1024 = 1.85449 \] You can try this if you have Excel, the formula is =LOG(number, base), so I typed in =LOG(1024,42). So to use the solution @malevolence19 gave you, you can solve for x I see what you did in your last response. The approach makes sense. My only question is, isn't is 42 to the exponent, or just 4 ???

OpenStudy (judygreeneyes):

Because 42 does not have a nice even exponent to get to 1024. I think have the problem as \[4^{2x+1}=1024\] makes a lot more sense at your level of math.

OpenStudy (anonymous):

Thank you so much!!!! 4 is just the number and 2 is the exponent. @judygreeneyes

OpenStudy (anonymous):

Well the 2x+1 is the exponent(s)

OpenStudy (judygreeneyes):

Then your approach is perfect: 2x+1 = 5, etc. Love it!

OpenStudy (anonymous):

Thank you so much you were a huge help!!!!!!!!!!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!