Solve the exponential equation. 900 = 1 0 x + 17 900=10^{x+17} 900 = 1 0 x + 17

In this problem, we're asked to solve the exponential equation. And here we're given the equation 900 is equal to 10 to the power of x plus 17. Now let's go ahead and get started with step 1 and isolate our exponential expression. But our exponential expression is actually already isolated. It's this 10 to the power of x plus 17, so we're good on step 1. We can move on to step 2. Now in step 2, we wanna check if there is a base 10 or not a base 10, and it looks like, yes, we do have a base 10, so we're going to just use log, the common log. So taking the log of both sides, I'm going to get log of 900 is equal to log of 10 to the power of x plus 17. Now that step 2 is done, we can move on to step 3 and use our log rules to get x out of the exponent. So in order to do that, I'm going to use the power rule and pull this x plus 17 to the front of my log. So this is log of 900 is equal to x plus 17 times log of 10. Now we can go ahead and simplify this further using log rules because we know that log of 10 is really just 1. So this goes away. It's just 1, and this is just log of 900 is equal to x plus 17. Now that we've completed step 3, we can go ahead and just solve for x. So solving for x here, I only have one thing to do and that's subtract this 17 from both sides. Now whenever you rewrite this, you need to be careful and use parentheses where needed or you can just put the 17 in at the front. It is up to you, but we just want to make sure that we are not making it seem like that 17 is being subtracted from the 900 because log 900 is something by itself and we're subtracting 17 from that. So this is a log of a 900. I'm gonna go ahead and use parentheses and then minus 17, and that's equal to x. Now if you're just using this in the log form, you can go ahead and stop here. But since we were done with step 4, we can go ahead and approximate this using a calculator. Now, plugging this into your calculator, make sure to use parentheses where appropriate because we're just taking the log of 900 and then subtracting 17 from that. Now, when you do that, you'll end up getting negative 14.05, rounding to that nearest hundredth place, and that is equal to x. And here is our final answer. Thanks for watching, and I'll see you in the next one.

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0. Functions4h 53m

Introduction to Functions
16m

Piecewise Functions
10m

Properties of Functions
9m

Common Functions
1h 8m

Transformations
5m

Combining Functions
27m

Exponent Rules
32m

Exponential Functions
28m

Logarithmic Functions
24m

Properties of Logarithms
34m

Exponential & Logarithmic Equations
35m

1. Limits and Continuity2h 2m

Introduction to Limits
41m

Finding Limits Algebraically
40m

Continuity
40m

2. Intro to Derivatives1h 33m

Tangent Lines and Derivatives
30m

Derivatives as Functions
14m

Basic Graphing of the Derivative
23m

Differentiability
26m

3. Techniques of Differentiation2h 18m

Basic Rules of Differentiation
43m

Higher Order Derivatives
12m

Product and Quotient Rules
55m

The Chain Rule
27m

4. Derivatives of Exponential & Logarithmic Functions1h 16m

Derivatives of Exponential & Logarithmic Functions
1h 16m

5. Applications of Derivatives2h 19m

Implicit Differentiation
27m

Logarithmic Differentiation
17m

Related Rates
59m

Differentials
21m

Linearization
13m

6. Graphical Applications of Derivatives6h 0m

Intro to Extrema
21m

The First Derivative Test
1h 20m

Concavity
1h 0m

The Second Derivative Test
22m

Curve Sketching
44m

Finding Global Extrema
29m

Applied Optimization
1h 41m

7. Antiderivatives & Indefinite Integrals48m

Antiderivatives
17m

Indefinite Integrals
31m

8. Definite Integrals4h 30m

Estimating Area with Finite Sums
42m

Riemann Sums
1h 16m

Introduction to Definite Integrals
20m

Fundamental Theorem of Calculus
25m

Average Value of a Function
17m

Substitution for Indefinite Integrals
55m

Substitution for Definite Integrals
32m

9. Graphical Applications of Integrals1h 43m

Area Between Curves
1h 19m

Introduction to Volume & Disk Method
23m

10. Integrals of Inverse, Exponential, & Logarithmic Functions21m

Integrals of Exponential Functions
14m

Integrals Involving Logarithmic Functions
7m

11. Techniques of Integration2h 7m

Integration by Parts
1h 37m

Improper Integrals
29m

12. Trigonometric Functions6h 54m

Angles
29m

Trigonometric Functions on Right Triangles
1h 8m

Solving Right Triangles
23m

Trigonometric Functions on the Unit Circle
1h 19m

Graphs of Sine & Cosine
46m

Graphs of Other Trigonometric Functions
32m

Trigonometric Identities
52m

Derivatives of Trig Functions
42m

Integrals of Basic Trig Functions
28m

Integrals of Other Trig Functions
10m

13: Intro to Differential Equations2h 23m

Basics of Differential Equations
43m

Slope Fields
19m

Separable Differential Equations
1h 21m

14. Sequences & Series2h 8m

Sequences
1h 12m

Review of Factorials
11m

Series
44m

15. Power Series2h 19m

Power Series & Taylor Series
2h 19m

16. Probability & Calculus45m

Continuous Probability Models
45m

0. Functions

Exponential & Logarithmic Equations

Business Calculus

0. Functions

Exponential & Logarithmic Equations

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Solve the exponential equation.
$900=10^{x+17}$

$x=-14.05$

$x=2.95$

$x=0.17$

$x=1.72$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Rewrite the given equation 900 = 10^(x+17) in logarithmic form to isolate the exponent. Using the property of logarithms log_b(a) = c if and only if b^c = a, we get log_10(900) = x + 17.

Simplify log_10(900) using a calculator or logarithmic rules. This will give you a numerical value for log_10(900).

Once you have the value of log_10(900), subtract 17 from both sides of the equation to isolate x. This gives x = log_10(900) - 17.

Evaluate the expression log_10(900) - 17 to find the value of x. Use a calculator to compute the final result.

Compare the computed value of x with the given answer choices to determine the correct one.