Q.Solve log8x+log4x+log2x=11.
Concept understanding — Laws of Logarithms
For a fixed base 0<a=1, the logarithmic function loga(⋅) is defined as the inverse of the exponential ax: y=ax⟺logay=x. Since ax has domain R and range (0,∞), loga has domain (0,∞) and range R -- logarithms of non-positive numbers are never defined. Since a0=1 always, loga1=0 for every base.
The five core laws (all provable directly from the inverse relationship):
alogax=x,loga(xy)=logax+logay,loga(yx)=logax−logay,logaxr=rlogax,logbx=logablogax.
Negative-base trick. log1/ax=−logax (immediate from the change-of-base formula with loga(1/a)=−1) -- very useful for combining mixed-base logarithmic equations into a single base.
Telescoping chains. A product like logab⋅logbc⋅logca always collapses to 1 (change every factor to a common base and watch every intermediate term cancel); more generally logab1⋅logb1b2⋯logbk−1bk=logabk.
Named bases. Base 10 = common logarithm; base e≈2.71828 (Euler's number, the limit of (1+1/n)n as n→∞, arising from continuous compounding) = natural logarithm, lnx; base 2 = binary logarithm, used in computer science.
Solving logarithmic equations. Convert to exponential form to remove the log entirely (e.g. log5−x(x2−6x+65)=2⇒(5−x)2=x2−6x+65), or combine several logs into one via the product/quotient/power rules first; ALWAYS check the resulting solution against the base condition (>0, =1) and the argument condition (>0) before accepting it.
Convert every term to base 2 using log2kx=k1log2x, then solve for log2x.
x=64.
Step 1. log8x=log28log2x=3log2x; log4x=2log2x; log2x=log2x.
Step 2. Let t=log2x. The equation becomes 3t+2t+t=11.
Step 3. Common denominator 6: 62t+63t+66t=611t=11⇒t=6.
Step 4. So log2x=6⇒x=26=64.
x=64.
Convert every logarithm to a common base, solve the resulting linear equation, then exponentiate
- Errors converting log8x to base 2 (dividing by log28=3, not multiplying).
- Forgetting to exponentiate at the end (leaving the answer as log2x=6).
- CBSE 2026Set ANNUAL1 markMCQQ.The value of log311⋅log1113⋅log1315⋅log1527 is:(a) 3(b) 1(c) 4(d) 2
›Reveal solutionSolution
The chained logarithms telescope via logab⋅logbc=logac, reducing the whole product to log327=3.
Using the identity logab=lnalnb, each factor can be written with natural logs:
log311⋅log1113⋅log1315⋅log1527=ln3ln11⋅ln11ln13⋅ln13ln15⋅ln15ln27
All the intermediate terms (ln11,ln13,ln15) cancel, leaving ln3ln27=log327.
Since 27=33, log327=3.
✓Final answerThe correct option is (a) 3.
- CBSE 2025Set ANNUAL1 markMCQQ.If 3 is the logarithm of 343, then the base is:(a) 6(b) 5(c) 9(d) 7
›Reveal solutionSolution
Convert the logarithmic statement to its exponential form and solve for the base.
"3 is the logarithm of 343 to base b" means logb343=3, i.e. b3=343.
Since 343=73, we get b3=73, so b=7.
✓Final answerThe correct option is (d) 7.
- CBSE 2024Set ANNUAL1 markMCQQ.The value of log2512 is:(a) 9(b) 16(c) 12(d) 18
›Reveal solutionSolution
log2512=18.
512=29 and 2=21/2.
log2512=log21/229=1/29=18,
using logak(am)=km.
✓Final answerlog2512=18 — option (d).
- CBSE 2022Set ANNUAL1 markMCQQ.The value of log2512 is:(a) 9(b) 16(c) 12(d) 18
›Reveal solutionSolution
log2512=18, found by expressing both numbers as powers of 2.
512=29 and 2=21/2.
log2512=log21/229=1/29=18, using the property logaman=mn.
✓Final answerThe correct option is (d) 18.
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