Question

log base 2(log x^5.4)-log base 2(log x^2.7)

Answer 1

we have to find the value of $\bf{log_2(logx^{5.4})-log_2(logx^{2.7})}$

solution : let's solve it step by step.

step 1 : solve $log_2(logx^{5.4})$ into simpler form

we know, logaⁿ = nloga

So, $log_2(logx^{5.4})=log_2(5.4logx)$

Step 2 : solve of $log_2(logx^{2.7})$ too, into simpler form.

i.e., $log_2(logx^{2.7})=log_2(2.7logx)$

step 3 : applying logarithmic rule, logm - logn = log(m/n)

so, $log_2(logx^{5.4})-log_2(logx^{2.7})$

= $log_2\left(\frac{5.4logx}{2.7logx}\right)$

= $log_2(2)$

We know, $log_a^a=1$

so, $log_22$ = 1

Therefore the value of $\bf{log_2(logx^{5.4})-log_2(logx^{2.7})}$ is 1