10(2^×) + 7(3^×) = 6^× + 70​

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

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.