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Exercise 2.12 · Q4

Q.Solve log⁡4(28x)=2log⁡28\log_4\left(2^{8x}\right)=2\log_2 8.

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Concept understanding — Laws of Logarithms

For a fixed base 0<a≠10<a\ne1, the logarithmic function log⁡a(⋅)\log_a(\cdot) is defined as the inverse of the exponential axa^x: y=ax  ⟺  log⁡ay=xy=a^x\iff\log_ay=x. Since axa^x has domain RR and range (0,∞)(0,\infty), log⁡a\log_a has domain (0,∞)(0,\infty) and range RR -- logarithms of non-positive numbers are never defined. Since a0=1a^0=1 always, log⁡a1=0\log_a1=0 for every base.

The five core laws (all provable directly from the inverse relationship):

alog⁡ax=x,log⁡a(xy)=log⁡ax+log⁡ay,log⁡a ⁣(xy)=log⁡ax−log⁡ay,log⁡axr=rlog⁡ax,log⁡bx=log⁡axlog⁡ab.a^{\log_ax}=x,\quad \log_a(xy)=\log_ax+\log_ay,\quad \log_a\!\left(\frac xy\right)=\log_ax-\log_ay,\quad \log_ax^r=r\log_ax,\quad \log_bx=\frac{\log_ax}{\log_ab}.

Negative-base trick. log⁡1/ax=−log⁡ax\log_{1/a}x=-\log_ax (immediate from the change-of-base formula with log⁡a(1/a)=−1\log_a(1/a)=-1) -- very useful for combining mixed-base logarithmic equations into a single base.

Telescoping chains. A product like log⁡ab⋅log⁡bc⋅log⁡ca\log_ab\cdot\log_bc\cdot\log_ca always collapses to 11 (change every factor to a common base and watch every intermediate term cancel); more generally log⁡ab1⋅log⁡b1b2⋯log⁡bk−1bk=log⁡abk\log_a b_1\cdot\log_{b_1}b_2\cdots\log_{b_{k-1}}b_k=\log_ab_k. …

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