1. The equation $$7^{x - 1} + 11^{x - 1} = 170$$ has
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By: anil on 05 May 2019 02.39 am
Let us start by putting the value of x.
When x = 1,
$$7^{x - 1} + 11^{x - 1} = 1+1 = 2$$
When x=2,
$$7^{x - 1} + 11^{x - 1} = 7+11 = 17$$
When x=3,
$$7^{x - 1} + 11^{x - 1} = 49+121 = 170$$
When x=4,
$$7^{x - 1} + 11^{x - 1} = 243+1331 = 1574$$
As we can see $$7^{x - 1} + 11^{x - 1}$$ is an increasing function and the only value that satisfies $$7^{x - 1} + 11^{x - 1} = 170$$ is x=3
Thus, the given equation has only 1 solution.
Hence, option B is the correct answer.
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When x = 1,
$$7^{x - 1} + 11^{x - 1} = 1+1 = 2$$
When x=2,
$$7^{x - 1} + 11^{x - 1} = 7+11 = 17$$
When x=3,
$$7^{x - 1} + 11^{x - 1} = 49+121 = 170$$
When x=4,
$$7^{x - 1} + 11^{x - 1} = 243+1331 = 1574$$
As we can see $$7^{x - 1} + 11^{x - 1}$$ is an increasing function and the only value that satisfies $$7^{x - 1} + 11^{x - 1} = 170$$ is x=3
Thus, the given equation has only 1 solution.
Hence, option B is the correct answer.