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burhan101

  • 3 years ago

Solve for x

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  1. burhan101
    • 3 years ago
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    \[\huge (5)(8)^{(x+2)} = 5^{7x}\]

  2. Hero
    • 3 years ago
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    \[5 \dot\ 8^{x+2} = 5^{7x} \\ 5 \dot\ 8^x 8^2 = \left(5{^7}\right)^x \\5 \dot\ 8^2 = \frac{\left(5{^7}\right)^x}{8^x} \\5 \dot\ 64 = \left(\frac{5^7}{8}\right)^x \\320 = \left(\frac{5^7}{8}\right)^x \] Take logs of both sides and finish simplifying

  3. burhan101
    • 3 years ago
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    |dw:1359086666173:dw|

  4. burhan101
    • 3 years ago
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    this is what i'm getting, but it is wrong

  5. burhan101
    • 3 years ago
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    wait, we're we sipposed to take the log of both sides?

  6. Hero
    • 3 years ago
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    \[\ln(320) = \ln\left(\left(\frac{5^7}{8}\right)^x\right) \\\ln(320) = x \ln\left(\frac{5^7}{8}\right) \\\frac{\ln(320)}{\ln\left(\frac{5^7}{8}\right)} = x \\\frac{\ln(320)}{\ln5^7 - \ln2^3} = x \\\frac{\ln(320)}{7\ln5 - 3\ln2} = x\]

  7. Hero
    • 3 years ago
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    Uh, ya....logs of both sides. What you do to one side, you do to the other side.

  8. Hero
    • 3 years ago
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    The last two lines, either form, is what you want.

  9. burhan101
    • 3 years ago
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    okay thanks !

  10. Hero
    • 3 years ago
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    By the way, don't forget that you are isolating x bro. Don't confuse the numerator and the denominator.

  11. burhan101
    • 3 years ago
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    yeah i know but im getting a weird answer idk why -.-

  12. Hero
    • 3 years ago
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    You should have gotten what I got above. And if you're approximating, you should get x ≈ 0.628

  13. burhan101
    • 3 years ago
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  14. Hero
    • 3 years ago
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    You did not isolate x properly bro.

  15. Hero
    • 3 years ago
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    Take a look at your second to last step.

  16. burhan101
    • 3 years ago
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    ohhh, i figured it. i flipped it

  17. burhan101
    • 3 years ago
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    got it ! thanks man :)

  18. Hero
    • 3 years ago
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    Wow....bro

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