Question

If f(x)= 9^x then prove that f(m+n+p) = f(m).f(n).f(p)

Answer 1

If f(x) = 7^x, then:

f(x + 3) = 7^(x + 3) and f(x + 1) = 7^(x + 1); therefore, ...

f(x + 3) - f(x + 1) = 7^(x + 3) - 7^(x + 1)

= (7^x)(7^3) - (7^x)(7^1) by one of the properties of exponents: The Product of Two Powers with the Same Base: (a^m)(a^n) = a^(m + n).

Now, factoring out a 7^x, we have:

= 7^x(7^3 - 7)

= 7^x(343 -7)

= 7^x(336)

= 336(7^x) is the final result.

Answer 2

${\huge \underline{\mathfrak\green{hy \:mate}}}$

$\huge\underline\bold \red {AnswEr : }$

If f(x) = 7^x, then:

f(x + 3) = 7^(x + 3) and f(x + 1) = 7^(x + 1); therefore, ...

f(x + 3) - f(x + 1) = 7^(x + 3) - 7^(x + 1)

= (7^x)(7^3) - (7^x)(7^1) by one of the properties of exponents: The Product of Two Powers with the Same Base: (a^m)(a^n) = a^(m + n).

Now, factoring out a 7^x, we have:

= 7^x(7^3 - 7)

= 7^x(343 -7)

= 7^x(336)

= 336(7^x) is the final result.

${\huge{\bf\blue{Thanks,}}}$