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Exercise 1.4 · Q1

Q.Expand log⁡b(aabbccdd)\log_b\left(\dfrac{a^{a} b^{b}}{c^{c} d^{d}}\right).

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61% · 11/18 Questions
✓ Free question

Split the quotient into a difference of logs, split each numerator/denominator product into a sum, then pull each exponent out front.

[!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;  log⁡bb=1\ \log_bb=1.

  1. Apply the quotient rule: log⁡b ⁣(aabbccdd)=log⁡b(aabb)−log⁡b(ccdd)\log_b\!\left(\dfrac{a^ab^b}{c^cd^d}\right)=\log_b(a^ab^b)-\log_b(c^cd^d).
  2. Apply the product rule to each: =[log⁡b(aa)+log⁡b(bb)]−[log⁡b(cc)+log⁡b(dd)]=\left[\log_b(a^a)+\log_b(b^b)\right]-\left[\log_b(c^c)+\log_b(d^d)\right].
  3. Apply the power rule to each term: =alog⁡ba+blog⁡bb−clog⁡bc−dlog⁡bd=a\log_ba+b\log_bb-c\log_bc-d\log_bd.
  4. Simplify log⁡bb=1\log_bb=1, so blog⁡bb=bb\log_bb=b: =alog⁡ba+b−clog⁡bc−dlog⁡bd=a\log_ba+b-c\log_bc-d\log_bd.
✓Final answer

log⁡b ⁣(aabbccdd)=alog⁡ba+b−clog⁡bc−dlog⁡bd\log_b\!\left(\dfrac{a^ab^b}{c^cd^d}\right)=a\log_ba+b-c\log_bc-d\log_bd.

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