evaluate 10^(log_10 (19))?

Question

the question is 10 to the power of log, base 10 to 19 if it's hard to read above. thanks

Answer 1

10^(log_10 (19))=19

Answer 2

Ten to the power of a log base 10 (log_10) will cancel itself out.

Answer 3

Although this is a difficult problem to typeset, it's remarkably straightforward to solve!

In fact, once you understand that the functions "10^" and "log(base 10)" are inverse functions -- that is, they "undo" each other in a mathematical sense -- then you would see the answer is ... 19!

If I were to "translate" the log and exponential functions using English, your problem might read like this:

"Log wants to know 'What power should I raise 10 to in order to make 19?' ... and then 10 wants to be raised to that exact power."

In other words, "Ten raised to the precise power of 10 that makes 19!" Which is ... ... 19!

Similarly, to stress my point about log and exp being inverse functions, consider this statement:

log(base10) (10^19) = 19

It works both ways, when functions are truly inverses of each other!

Hope this helps!

Answer 4

It's 19. 10^(log_10 x) is always x.

Answer 5

10^(log_10(19)) = 19 by inverse functions of common logarithms and exponencial function 10^ x

10^(log_10(x)) = x

Answer 6

The log function is the inverse of the power function. This is the same as asking for the square of the square root of 19, or 1/(1/19), or -(-19) or (19*5362)/5362 ...

Answer 7

Whenever you have something of the form

b^(log[base b])(x)

The b^ and the log[base b] will cancel each other out because they are inverses of each other.

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Proof: Let y = b^(log [base b](x)) Changing this to logarithmic form, we have

log[base b](y) = log[base b](x), therefore
y = x.
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In our case, 10^(log[base 10](19)) = 19.