pushpitkc wrote:
dave13 wrote:
Hi
pushpitkc, can you please explain some steps below ?
\(2^x-2^{x}*2^{-2}=3*2^{13}\)
how do we additional 2 on the left side ? \(2^x-\frac{2^{x}}{2^{2}}=3*2^{13}\)
how do we get fraction on the left side ? \(2^x(1-\frac{1}{2^{2}})=3*2^{13}\)
how after this \(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\)
we get this on the left side \(2^x*\frac{3}{2^{2}}=3*2^{13}\)
and how we get this on the left side ? \(2^x=2^{15}\)
hopefully this step will be clear after the previous one \(x=15\).
many thanks
Hi
dave13Some formulae in exponents that you must remember in exponents
\(a^0 = 1\) | \(a^1 = a\) | \(\sqrt{a} = a^{\frac{1}{2}}\) | \(a^{-n} = \frac{1}{a^n}\)
\(a^{m+n} = a^m*a^n\) | \(\frac{a^m}{a^n}=a^{m−n}\) | \(a^n = \frac{1}{a^{-n}}\)
Coming back to the problem at hand
\(2^x-2^{x}*2^{-2}=3*2^{13}\)
\(2^x-\frac{2^{x}}{2^{2}}=3*2^{13}\)
Here, \(2^{x-2} = 2^{x + (-2)} = 2^x*2^{-2}\) \(2^x(1-\frac{1}{2^{2}})=3*2^{13}\)
Here, we take \(2^x\) as common from both the expressions\(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\)
Here, we simplify \(1 - \frac{1}{2^2}\) as \(\frac{2^2 - 1}{2^2} = \frac{4-1}{2^2} = \frac{3}{2^2}\)\(2^x*\frac{3}{2^{2}}=3*2^{13}\) -> \(2^x=2^{15}\)
Here, we take \(2^2\) from left hand side to right hand side pushpitkc,
thanks for explanation, i somehow dont get based on which formula did you solve the below expression, (you provide all formulas but i couldnt recognize the pattern (i mean which formula to apply)
Here, \(2^{x-2} = 2^{x + (-2)} = 2^x*2^{-2}\) can you please explain somehow
thanks and have a good weekend
generis perhaps you can explain from different angle if such angle exists ?
I chase quant like dog chases its tale …
trying to "catch" understanding ..., having hard time dealing with exponents now
hope are enjoying the weekend too