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3 years ago

10(2^×) + 7(3^×) = 6^× + 70

Mathematics

1 answer:

Mkey [24]3 years ago

8 0

Answer:

x = 2.81 and 2.096

Step-by-step explanation:

Given the expression

10(2^x) + 7(3^x) = 6^x + 70

This can also be expressed as;

10(2^x) + 7(3^x) = (2*3)^x + 70

10(2^x) + 7(3^x) = 2^x*3^x + 70

Let a = 2^x and b = 3^x

10a + 7b = ab + 70

10a + 7b - ab = 70

10a-ab + 7b - 70 = 0

a(10-b)+7(b-10) = 0

a(10-b)-7(10-b) = 0

a-7 = 0 and 10-b = 0

a = 7 and b = 10

Since a = 2^x

7 = 2^x

log 7 = log2^x

log7 = xlog2

x = log7/log2

x = 2.81

Similarly

10 = 3^x

log 10 = log 3^x

log 10 = xlog3

x = log 10/log 3

x = 1/0.4771

x = 2.096

Hence the values of x that satisfies the equation are 2.81 and 2.096

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3 years ago

10(2^×) + 7(3^×) = 6^× + 70​

10(2^×) + 7(3^×) = 6^× + 70