Question

2020^(1+log2 to the base 2020)

Answer 1

Given,

$\bf{2020^{(1 + log_{2020}2)}}$

we have to find the value of given expression.

step 1 : first solve the $(1 + log_{2020}2)$

we know, $log_aa=1$

so, $log_{2020}2020=1$

now $(1+log_{2020}2)=log_{2020}2020+log_{2020}2$

now using concept,$log_am+log_an=log_a(mn)$

so, $log_{2020}2020+log_{2020}2=log_{2020}(2020\times2)$

= $log_{2020}4040$

now $\bf{2020^{(1 + log_{2020}2)}}$ converts into $2020^{log_{2020}4040}$

we know, $a^{log_ab}=b$

so, $2020^{log_{2020}4040=4040$

therefore value of given logarithmic expression is 4040.

Answer 2

Answer:hello & namasthe if u r satisfied with this explanation then pls mark me as brainliest as a favour to me by u.nice to answer u

Given

Evaluate 2020^(1+log 2 base 2020)

Step-by-step explanation:

Firstly we have to find the value of (1+log 2base 2020)

By solving we get as follows

1 in (1+log 2base 2020) can be written as log 2020 base 2020 right!!!

Becoz As we know log x to base x=1

so, substitute 1 as log 2020 base 2020 in (1+log 2base 2020) now it is converted into

(1+log 2base 2020)--->(log 2020 base 2020 +log 2 base 2020)

As we know, log x +y base z=log xy base z right!!!

so, (log 2020 base 2020 +log 2 base 2020)=

(log 2020 × 2 base 2020)

=(log 4040 base 2020)

1+log 2 base 2020)--->(log 4040 base 2020)

Now substitute this in 2020^(1+log 2 base 2020)

2020^(log 4040 base 2020)

As we know log a^(log m base x) = m right!!!

4040

.•.so the value of given logarithmic expression is 4040.