### 6 Answers

Relevance

- PinkgreenLv 71 month ago
Using the calculator directly, we have

y=0.851677177 approximately.

or

log(10)y=log(10)0.65+1.03log(10)1.3

=>

log(10)y=-0.18708664+1.03(0.113943352)

=>

log(10)y=-0.06972499

=>

y=0.851677178 (anti log)

or

y=10^(-0.06972499)=0.851677178.

- Login to reply the answers

- PhilipLv 61 month ago
y = 0.65(1.3)^(1.03). lny =ln(0.65)+(1.03)ln(1.3)

= -4.307829161 + (1.03)(0.2623642645) =

-1.605477237. Then y=e^lny=e^(-1.605477237),

ie., y = 0.851677177.

- Login to reply the answers

- How do you think about the answers? You can sign in to vote the answer.

Still have questions? Get your answers by asking now.