Question

2.
log (2500) on bases of 5 root 2
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Answer 1

The logarithm function with base b of a number x is written as$log_b(x$) and it is the exponent to which we must raise the base b to get the number x.

In other words,

$log_b(x) = y = > b^y = x$

To solve for $log_b(x) = y,$ we can use the following rule

$log_b(a^n) = n * log_b(a)$

So to solve for $log_5^22 (2500$), we can use the following rule:

$log_5$√2(2500)

= $log_5$√2$(5^4 * 2^4)$

=$4 * log_5$√2$(5) + 4 * log_5$√2(2)

Now we can simplify the above expression:

$log_5$√2(2500)

= $4 * log_5$√2(5) + $4 * log_5$√2(2)

= $4 * (log_5$(5) + log_√2(2))

= 4 * (1 + 0.5) = 4*1.5

= 6

So the solution is$log_5$√2 (2500) = 6

This means that $5^\sqrt2^6 = 2500$

Know more from the following links.

brainly.in/question/16766451

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