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For a real number aa, if log⁡15a+log⁡32a(log⁡15a)(log⁡32a)=4\frac{\log _{15} a+\log _{32} a}{\left(\log _{15} a\right)\left(\log _{32} a\right)}=4 then a must lie in the range.

Solution

✅ Correct Option: 3

We can see the reference solution jumps through several steps without proper explanation. Let us break this down so you can follow every move and understand the key concepts.


The key insight here is using the change of base formula: log⁡bx=log⁡xlog⁡b\log_b x = \frac{\log x}{\log b}

This lets us convert any logarithm to natural logs (or any common base).

So our given terms become:

log⁡15a=log⁡alog⁡15\log_{15} a = \frac{\log a}{\log 15}

log⁡32a=log⁡alog⁡32\log_{32} a = \frac{\log a}{\log 32}


Our equation: log⁡15a+log⁡32a(log⁡15a)(log⁡32a)=4\frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = 4

Becomes: log⁡alog⁡15+log⁡alog⁡32log⁡alog⁡15×log⁡alog⁡32=4\frac{\frac{\log a}{\log 15} + \frac{\log a}{\log 32}}{\frac{\log a}{\log 15} \times \frac{\log a}{\log 32}} = 4


log⁡alog⁡15+log⁡alog⁡32=log⁡a(1log⁡15+1log⁡32)\frac{\log a}{\log 15} + \frac{\log a}{\log 32} = \log a \left(\frac{1}{\log 15} + \frac{1}{\log 32}\right)

Using the common denominator rule:

=log⁡a⋅log⁡32+log⁡15log⁡15⋅log⁡32= \log a \cdot \frac{\log 32 + \log 15}{\log 15 \cdot \log 32}


log⁡alog⁡15×log⁡alog⁡32=(log⁡a)2log⁡15⋅log⁡32\frac{\log a}{\log 15} \times \frac{\log a}{\log 32} = \frac{(\log a)^2}{\log 15 \cdot \log 32}


Our equation now looks like:

log⁡a⋅log⁡32+log⁡15log⁡15⋅log⁡32(log⁡a)2log⁡15⋅log⁡32=4\frac{\log a \cdot \frac{\log 32 + \log 15}{\log 15 \cdot \log 32}}{\frac{(\log a)^2}{\log 15 \cdot \log 32}} = 4

Here's the magic: When we divide fractions, we multiply by the reciprocal:

log⁡a⋅(log⁡32+log⁡15)(log⁡a)2=4\frac{\log a \cdot (\log 32 + \log 15)}{(\log a)^2} = 4

Notice how the (log⁡15⋅log⁡32)(\log 15 \cdot \log 32) terms cancel out!


log⁡32+log⁡15log⁡a=4\frac{\log 32 + \log 15}{\log a} = 4

Cross-multiplying: log⁡32+log⁡15=4log⁡a\log 32 + \log 15 = 4 \log a


Key Property: log⁡x+log⁡y=log⁡(xy)\log x + \log y = \log(xy) and nlog⁡x=log⁡xnn \log x = \log x^n

So: log⁡(32×15)=log⁡a4\log(32 \times 15) = \log a^4

log⁡480=log⁡a4\log 480 = \log a^4

Therefore: a4=480a^4 = 480


We need a=4804a = \sqrt[4]{480}

Let's use perfect fourth powers to estimate:

44=2564^4 = 256

54=6255^4 = 625

Since 256<480<625256 < 480 < 625, we have:

4<4804<54 < \sqrt[4]{480} < 5

Therefore: 4<a<54 < a < 5


Key Takeaway: The change of base formula is your best friend when dealing with logarithms of different bases. Always look for opportunities to simplify complex logarithmic expressions by factoring out common terms!

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