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abdulrehman95a

  • 4 years ago

What is the solution to 3^4x-7= 4^2x+3

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  1. rivermaker
    • 4 years ago
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    Is it \[3^{4}x - 7 = 4 ^{2}x + 3\] or \[3^{4x} - 7 = 4 ^{2x} + 3\]

  2. .Sam.
    • 4 years ago
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    or \[3^{4x-7}=4^{2x+3}\]

  3. abdulrehman95a
    • 4 years ago
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    Sam's right

  4. mpt003
    • 4 years ago
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    10/65

  5. abdulrehman95a
    • 4 years ago
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    explanations please?

  6. rivermaker
    • 4 years ago
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    \[3^{4x-7}=4^{2x+3}\] \[\frac{3^{4x}}{3^{7}}=4^{2x}\times {4^3}\] \[\frac{81^x}{2187} = 64\times16^x\] \[(\frac{81}{16})^x = 64 \times 2187\] Now takes logs on both sides

  7. rivermaker
    • 4 years ago
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    \[x = \frac{(\log 64+\log2187)}{\log 81-\log 16}\]

  8. abdulrehman95a
    • 4 years ago
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    wow thanks!

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