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|dw:1359089106003:dw|
i can substitute 2^2 for u for all terms but the last .. idk what to do !
hint: first put them all to the lowest base of 2.

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\[\frac{ 1 }{ \sqrt{2} }=2^{-\frac{ 1 }{ 2 }}\]
yep, @Azteck has shown u the harder one, try to do the rest yourself. Post it on here so I/others can help u check it.
@JayDS He only wanted to know the last one. Read what he said above.
oh kk sorry, I didn't quite understand how he phrased it.
\[4^2(2^{x - 3}) = 16^{x-2}\] Is that the whole thing @burhan101 ?
Because if it is, it works out quite nicely
no @Hero you are missing the last term
|dw:1359091740887:dw|
\[4^2 \dot\ 2^{x - 3} = \frac{16^{x - 2}}{\sqrt{2}} \\4^2\dot\ 2^x2^{-3} = \frac{16^x16^{-2}}{\sqrt{2}} \\\frac{4^2 \dot\ 2^x}{2^3} = \frac{16^x}{16^2 \sqrt{2}} \\\frac{4^2 \dot\ 16^2 \sqrt{2}}{2^3}= \frac{16^x}{2^x} \\512\sqrt{2} = \left(\frac{16}{2}\right)^x \\512\sqrt{2} = \left(8\right)^x \] You should be able to finish it from there by taking logs of both sides and simplifying
thank you but this is not the method my teacher wants, she wants us to substitute
Substitute what?
like a variable like let's say use q to represent something
Why would she want you to do that? It simplifies quite nicely without it.
I wish i knew -.- something about knowing a wide variety of methods
If you take logs of both sides you get: \[\ln(512 \dot\ \sqrt{2}) = x \ln(8) \\\frac{\ln(512) + \ln(\sqrt{2})}{\ln(8)} = x \\\frac{\ln(2^9) + \ln(2^{1/2})}{\ln(2^3)} = x \\\frac{9\ln(2) + .5\ln(2)}{3\ln(2)} = x \\\frac{9.5 \ln(2)}{3 \ln(2)} = x \\\frac{9.5}{3} = x \]
\[(2^2)^2(2^{2x-3})=(16^{x-2})(\frac{ 1 }{ \sqrt{2} })\] \[(2^4)(2^{2x-3})=(2^{4x-8})(2^{-\frac{ 1 }{ 2 }})\] \[2^{2x-3+4}=2^{4x-8-\frac{ 1 }{ 2 }}\] \[2^{2x+1}=2^{4x-\frac{ 17 }{ 2 }}\] \[2x+1=4x-\frac{ 17 }{ 2 }\] \[4x+2=8x-17\] Solve for x.
@Hero you copied the wrong indice.
Good job bro.

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