10-10-2016
Mathematics

contestada

Solve for x. what is 1900=1300(1+.0625)^x?

Respuesta :

10-10-2016

First rearrange as:
[tex]\frac{1900}{1300}=1.0625^{x}[/tex]
Taking logs of both sides, we get:
[tex]log\ \frac{1900}{1300}=x\ log1.0625[/tex]
[tex]x=\frac{log\frac{1900}{1300}}{log\ 1.0625}=6.26[/tex]
x = 6.26.

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Panoyin Panoyin

10-10-2016

1900 = 1300(1 + 0.0625)ˣ
1900 = 1300(1.0625)ˣ
1300 1300
1⁶/₁₃ = (1.0625)ˣ
㏑(1⁶/₁₃) = ln((1.0625)ˣ)
㏑(1⁶/₁₃) = x㏑(1¹/₁₆)
ln(1¹/₁₆) ln(1¹/₁₆)
6.26 ≈ x

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