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Exercise 1.4 · Q2

Q.Expand log⁡b(4x69y7)\log_b\left(\dfrac{4x^{6}}{9y^{7}}\right).

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✓ Free question

Split the quotient into a difference, the numerator/denominator products into sums, then move each power out front (and 4=224=2^2, 9=329=3^2 can be reduced further).

[!FORMULA]

log⁡b ⁣(MN)=log⁡bM−log⁡bN\log_b\!\left(\dfrac{M}{N}\right)=\log_bM-\log_bN;  log⁡b(MN)=log⁡bM+log⁡bN\ \log_b(MN)=\log_bM+\log_bN;  log⁡b(Mk)=klog⁡bM\ \log_b(M^{k})=k\log_bM.

  1. Quotient rule: log⁡b ⁣(4x69y7)=log⁡b(4x6)−log⁡b(9y7)\log_b\!\left(\dfrac{4x^6}{9y^7}\right)=\log_b(4x^6)-\log_b(9y^7).
  2. Product rule on each: =[log⁡b4+log⁡bx6]−[log⁡b9+log⁡by7]=\left[\log_b4+\log_b x^6\right]-\left[\log_b9+\log_by^7\right].
  3. Power rule on the variable terms: =log⁡b4+6log⁡bx−log⁡b9−7log⁡by=\log_b4+6\log_bx-\log_b9-7\log_by.
  4. Optionally reduce the constants (4=224=2^2, 9=329=3^2) via the power rule: log⁡b4=2log⁡b2\log_b4=2\log_b2, log⁡b9=2log⁡b3\log_b9=2\log_b3, giving =2log⁡b2+6log⁡bx−2log⁡b3−7log⁡by=2\log_b2+6\log_bx-2\log_b3-7\log_by.
✓Final answer

log⁡b ⁣(4x69y7)=log⁡b4+6log⁡bx−log⁡b9−7log⁡by=2log⁡b2+6log⁡bx−2log⁡b3−7log⁡by\log_b\!\left(\dfrac{4x^6}{9y^7}\right)=\log_b4+6\log_bx-\log_b9-7\log_by=2\log_b2+6\log_bx-2\log_b3-7\log_by.

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