## burhan101 Group Title Solve one year ago one year ago

1. burhan101 Group Title

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2. burhan101 Group Title

i can substitute 2^2 for u for all terms but the last .. idk what to do !

3. JayDS Group Title

hint: first put them all to the lowest base of 2.

4. Azteck Group Title

$\frac{ 1 }{ \sqrt{2} }=2^{-\frac{ 1 }{ 2 }}$

5. JayDS Group Title

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.

6. Azteck Group Title

@JayDS He only wanted to know the last one. Read what he said above.

7. JayDS Group Title

oh kk sorry, I didn't quite understand how he phrased it.

8. Hero Group Title

$4^2(2^{x - 3}) = 16^{x-2}$ Is that the whole thing @burhan101 ?

9. Hero Group Title

Because if it is, it works out quite nicely

10. burhan101 Group Title

no @Hero you are missing the last term

11. burhan101 Group Title

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12. Hero Group Title

$4^2 \dot\ 2^{x - 3} = \frac{16^{x - 2}}{\sqrt{2}} \\4^2\dot\ 2^x2^{-3} = \frac{16^x16^{-2}}{\sqrt{2}} \\\frac{4^2 \dot\ 2^x}{2^3} = \frac{16^x}{16^2 \sqrt{2}} \\\frac{4^2 \dot\ 16^2 \sqrt{2}}{2^3}= \frac{16^x}{2^x} \\512\sqrt{2} = \left(\frac{16}{2}\right)^x \\512\sqrt{2} = \left(8\right)^x$ You should be able to finish it from there by taking logs of both sides and simplifying

13. burhan101 Group Title

thank you but this is not the method my teacher wants, she wants us to substitute

14. Hero Group Title

Substitute what?

15. burhan101 Group Title

like a variable like let's say use q to represent something

16. Hero Group Title

Why would she want you to do that? It simplifies quite nicely without it.

17. burhan101 Group Title

I wish i knew -.- something about knowing a wide variety of methods

18. Hero Group Title

If you take logs of both sides you get: $\ln(512 \dot\ \sqrt{2}) = x \ln(8) \\\frac{\ln(512) + \ln(\sqrt{2})}{\ln(8)} = x \\\frac{\ln(2^9) + \ln(2^{1/2})}{\ln(2^3)} = x \\\frac{9\ln(2) + .5\ln(2)}{3\ln(2)} = x \\\frac{9.5 \ln(2)}{3 \ln(2)} = x \\\frac{9.5}{3} = x$

19. Azteck Group Title

$(2^2)^2(2^{2x-3})=(16^{x-2})(\frac{ 1 }{ \sqrt{2} })$ $(2^4)(2^{2x-3})=(2^{4x-8})(2^{-\frac{ 1 }{ 2 }})$ $2^{2x-3+4}=2^{4x-8-\frac{ 1 }{ 2 }}$ $2^{2x+1}=2^{4x-\frac{ 17 }{ 2 }}$ $2x+1=4x-\frac{ 17 }{ 2 }$ $4x+2=8x-17$ Solve for x.

20. Azteck Group Title

@Hero you copied the wrong indice.

21. Hero Group Title

Good job bro.