Question

log 625 base 4/ log 343 base 243 -log 625 base 8/ log 49 base 243=​

Answer 1

The answer is zero attached below is how to solve the problem. I am really sorry about my handwriting as it looks like chicken scratch

Answer 2

Concept:

Change of base rule of logarithms.

logₐ b where a is base.

Now according to the change of base rule, it can be written as,

logₐ b = (log₁₀ b)/(log₁₀ a) or $log_e b/log_e a$

Given:

log 625 base 4/ log 343 base 243 -log 625 base 8/ log 49 base 243

log₄625/log₂₄₃343 - log₈625/log₂₄₃49

Find:

The value of the given equation.

Solution:

= log₄625/log₂₄₃343 - log₈625/log₂₄₃49

= $[\frac{log_{10} 625}{log_{10} 4} \ \div \ \frac{log_{10} 343}{log_{10} 243}] - [\frac{log_{10} 625}{log_{10} 8} \ \div \ \frac{log_{10} 49}{log_{10} 243} ]$

= $[\frac{log_{10} 625}{log_{10} 4} \ \times \ \frac{log_{10} 243}{log_{10} 343}] - [\frac{log_{10} 625}{log_{10} 8} \ \times \ \frac{log_{10} 243}{log_{10} 49} ]$

= $[\frac{log_{10} 5^4}{log_{10} 2^2} \ \times \ \frac{log_{10} 3^5}{log_{10} 7^3}] - [\frac{log_{10} 5^4}{log_{10} 2^3} \ \times \ \frac{log_{10} 3^5}{log_{10} 7^2} ]$

= $[\frac{4 \ log_{10} 5}{2 \ log_{10} 2} \ \times \ \frac{5 \ log_{10} 3}{3 \ log_{10} 7}] - [\frac{4 \ log_{10} 5}{3 \ log_{10} 2} \ \times \ \frac{5 \ log_{10} 3}{2 \ log_{10} 7} ]$        [Using log mⁿ = n log m]

= $\frac{20 \ log_{10} 5 \ log_{10} 3}{6 \ log_{10} 2 \ log_{10} 7 } - \frac{20 \ log_{10} 5 \ log_{10} 3}{6 \ log_{10} 2 \ log_{10} 7 }$

= 0

Hence, the value of the above equation is 0.

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