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Mathematics 14 Online
OpenStudy (anonymous):

Is Anyone Here Good With Logarithms? I Really Could Use Some Help On A HomeWork Thing. Please

OpenStudy (anonymous):

2^x+1=8x

OpenStudy (anonymous):

is it (2^x) +1 = 8x

OpenStudy (anonymous):

No Its 2^(x+1)=8x

zepdrix (zepdrix):

It's not 8^x on the right side? You sure? :U

zepdrix (zepdrix):

\[\Large\rm 2^{x+1}=8^x\]Cuz this problem would make a lot more sense :)

zepdrix (zepdrix):

8 is a power of 2, more specifically, it's 2 to the 3rd power.\[\Large\rm 2^{x+1}=8^x\qquad\to\qquad 2^{x+1}=\left(2^{3}\right)^x\]Applying rules of exponents gives us:\[\Large\rm 2^{x+1}=2^{3x}\]Since the bases are the same, the exponents have to be equal as well. So equate the exponents and solve for x!

OpenStudy (anonymous):

That's It!

OpenStudy (anonymous):

Wait How Do You Equate Exponents? Do You Mean Like Put Them Together?

OpenStudy (amistre64):

if you cant tell of one thing is an base of the other, just log it out ....

zepdrix (zepdrix):

Equate! :)\[\Large\rm x+1=3x\]Setting the exponents equal to one another. Understand you're only allowed to do this since the bases are equal!

OpenStudy (anonymous):

Oh Okay So Its 1=2x?

zepdrix (zepdrix):

Mmmm yah that would be ...a... step +_+ finish it up though!

OpenStudy (amistre64):

\[2^{x+1}=8^x\] \[log(2^{x+1})=log(8^x)\] \[(x+1)log(2)=x~log(8)\] \[x~log(2)+log(2)=x~log(8)\] \[log(2)=x~log(8)-x~log(2)\] \[log(2)=x~(log(8)-log(2))\] \[\frac{log(2)}{log(8)-log(2)}=x\] etc ...

OpenStudy (anonymous):

x=1/2? Which Would Mean 2^1/2 or x=1/2

zepdrix (zepdrix):

yay good job \c:/

OpenStudy (anonymous):

That's It? Will You Please Help Me With ANother One? Logarithims Are So Confusing!

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