Is Anyone Here Good With Logarithms? I Really Could Use Some Help On A HomeWork Thing. Please
2^x+1=8x
is it (2^x) +1 = 8x
No Its 2^(x+1)=8x
It's not 8^x on the right side? You sure? :U
\[\Large\rm 2^{x+1}=8^x\]Cuz this problem would make a lot more sense :)
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!
That's It!
Wait How Do You Equate Exponents? Do You Mean Like Put Them Together?
if you cant tell of one thing is an base of the other, just log it out ....
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!
Oh Okay So Its 1=2x?
Mmmm yah that would be ...a... step +_+ finish it up though!
\[2^{x+1}=8^x\] \[log(2^{x+1})=log(8^x)\] \[(x+1)log(2)=x~log(8)\] \[x~log(2)+log(2)=x~log(8)\] \[log(2)=x~log(8)-x~log(2)\] \[log(2)=x~(log(8)-log(2))\] \[\frac{log(2)}{log(8)-log(2)}=x\] etc ...
x=1/2? Which Would Mean 2^1/2 or x=1/2
yay good job \c:/
That's It? Will You Please Help Me With ANother One? Logarithims Are So Confusing!
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