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How will the graph of log x compare to the graph of ln x?

User Redjamjar
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logx=log_10(x), ln(x)=log_e(x). These two graphs have the same y-intercept (1, 0), because log(1)=ln(1)=0 as any log function is. However, when x>1, ln(x) will rise more quickly. For example, when x=e, ln(x) is 1. However, for log(x) to be 1, x has to be 10, and 10>e. The smaller the base is, the more quickly the function increases. 
User Daksh Gargas
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Answer:

Explanation:

Logx means log of base 10

And In(x) means natural logarithm of base e

When will log and In(x) be equal

They are equal at x = 1,

So, log1 = 0 and In(1) = 0

So they intercept at x = 1

Also, it is notice from the graph that the log of their negative numbers are undefined as x<-1 for Logx and x<-3 for In(x)

Check attachment to compare the graphs

It is notice that

In(x) is increasing at a faster rate as x increases

Logx is stretched more as x increases.

So, the major difference between them is that the natural logarithm In(x) increases at a faster rate than the common logarithmic (Logx)

The red line indicates In(x)

The blue line indicates Logx

How will the graph of log x compare to the graph of ln x?-example-1
User Thomas Turner
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