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If $f(x)=10^{x},$ show that $\frac{f(x+h)-f(x)}{h}=10^{x}\left(\frac{10^{h}-1}{h}\right)$

$10^{x} \cdot \frac{10^{h}-1}{h}$

Algebra

Chapter 5

Exponential and Logarithmic Functions

Section 1

Exponential Functions

Missouri State University

Idaho State University

Lectures

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here. We have ever backs being equal to 10 to the X. So now you want to use on a difference question here. So f off X plus h minus after vax all divided by H. Ok, and this is supposed to be equal to 10 to the ex. Um right. Wanna show that this is equal to 10 to the X times 10 to the age, minus one all over each. Okay, let's go ahead and just plug this end to a different question. So what is f of X plus h mean while f of axe is 10 to the axe, so f of X plus h would be 10 to the X plus a h. Ok, there's f of X plus age and then we have one minus aftereffects. So minus f of X is minus. Whatever backs is 10 to the X. We have minus 10 to the axe, right? I'm all divided by h. So there we have. And this is f of X plus age minus half of axe, all divided by age. Now, this is supposed be equal to this. It is not yet, but what? What What What could we do here? Well, we can go ahead and actually factor out a 10 to the X from the numerator. If we do that, we get in fact, er out of 10 to the ax. Okay, Now we'll tend to the x Times. What? Give me a 10 to the X plus h while will be 10 to the h. Why? Because we multiply right with the same base. We add the exponents. So therefore, a 10 to the axe. Times 10 to the H. Well, that's equal to 10 to the X plus a h or then right. I'm factoring out a 10 to the X. So 10 to the X Times. What gives me a negative 10 to the X. Well, that's minus one. Okay, so there's my numerator factor. Right? But then that is all over all over h. Okay, which well, that tend to the X right can just come outside and we get, um this is equal to 10 to the X times 10 to the H minus one over h and therefore right. We prove that is the expression we want, and we got it. So therefore, um, we have shown, um What? Want to show? All right, take care

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