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1) Explain what happens to the graph of the function f(x) = log(b)x when 'b' is equal...

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1) Explain what happens to the graph of the function f(x) = log(b)x when 'b' is equal to 1?

2) Is the following statment true or false, when 'b' = 1, g(x) = b^x and h(x) = log(b)x intersect at the ordered pair of (10,1)?

Remember, (b) - represents base which is a subscript to log.

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Posted (Answer #1)

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We want to examine the graph of: `y = log_1 x` . 

If we transform this to its exponential form, we get:

`1^y = x`

Hence, `y = log_1 x` cannot be a function of `x` because you cannot satisfy the relation for any value of `x ne 1` . If you try to plot this, using points that satisfy the relation, you will get a straight, vertical line, `x = 1` - since `1^y` is always equal to 1 for any value of `y` . Again, the given expression is not a function.

For the second question, we want to know if, when our base is 1, `g(x) = b^x` and` h(x) = log_b x` [or equivalently `g(x) = 1^x ` and `h(x) = log_1 x ` ] intersect at `(10, 1)` . 

Note that 1 raised to any power is just 1. Due to this property, g(x) will just be a horizontal line, `y=1` , as it can only be equal to 1 for any value of x. Then, as mentioned earlier, h(x) [though not a valid function; also for the reason that it's graph, if plotted, won't pass the vertical line test], will result to a vertical line `x=1` .

Hence, the intersection will not be at (10, 1). The graphs intersect at (1, 1) -- the point where x = y = 1. 

[Please check your question. If question 2 is not a typo, then the answer is false because the intersection will be at (1, 1) NOT (10, 1). If it's a typo and (10, 1) should have been (1, 1), then the answer is true]

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Posted (Reply #1)

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Thank you for your detailed response. Yes, the point of (10,1) was supposed to be (1,1) - just added an unneeded zero in the x-value position, definitely a typo!

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