Exponential Growth and Decay Practice

Start from 100%, then add or subtract the stated percentage before converting to a decimal.

Convert the rate to \(0.06\). A growth factor is \(1+r=1.06\).

Start from 100%, then add or subtract the stated percentage before converting to a decimal.

Convert the rate to \(0.18\). A decay factor is \(1-r=0.82\).

Start from 100%, then add or subtract the stated percentage before converting to a decimal.

Convert the rate to \(0.125\). A growth factor is \(1+r=1.125\).

Start from 100%, then add or subtract the stated percentage before converting to a decimal.

Convert the rate to \(0.04\). A decay factor is \(1-r=0.96\).

Start from 100%, then add or subtract the stated percentage before converting to a decimal.

Convert the rate to \(0.35\). A growth factor is \(1+r=1.35\).

Identify P, convert the rate, choose 1 + r or 1 - r, and make the number of periods the exponent.

The initial value is \(P=2500\). The growth multiplier is \(1.07\), so \(A=2500(1.07)^t\); it returns the correct intercept and plotted values.

Identify P, convert the rate, choose 1 + r or 1 - r, and make the number of periods the exponent.

The initial value is \(P=32000\). The decay multiplier is \(0.85\), so \(A=32000(0.85)^t\); it returns the correct intercept and plotted values.

Identify P, convert the rate, choose 1 + r or 1 - r, and make the number of periods the exponent.

The initial value is \(P=180\). The growth multiplier is \(1.12\), so \(A=180(1.12)^t\); it returns the correct intercept and plotted values.

Identify P, convert the rate, choose 1 + r or 1 - r, and make the number of periods the exponent.

The initial value is \(P=900\). The decay multiplier is \(0.92\), so \(A=900(0.92)^t\); it returns the correct intercept and plotted values.

Identify P, convert the rate, choose 1 + r or 1 - r, and make the number of periods the exponent.

The initial value is \(P=6400\). The growth multiplier is \(1.045\), so \(A=6400(1.045)^t\); it returns the correct intercept and plotted values.

Evaluate the exponent before multiplying and round only the final calculator result.

Use the full growth factor: \(A=1200(1.05)^{3}=1389.15\). The graph's labeled endpoint is generated from this value.

Evaluate the exponent before multiplying and round only the final calculator result.

Use the full growth factor: \(A=450(1.08)^{4}=612.22\). The graph's labeled endpoint is generated from this value.

Evaluate the exponent before multiplying and round only the final calculator result.

Use the full growth factor: \(A=8000(1.025)^{2}=8405\). The graph's labeled endpoint is generated from this value.

Evaluate the exponent before multiplying and round only the final calculator result.

Use the full growth factor: \(A=75(1.2)^{3}=129.6\). The graph's labeled endpoint is generated from this value.

Evaluate the exponent before multiplying and round only the final calculator result.

Use the full growth factor: \(A=3200(1.06)^{5}=4282.32\). The graph's labeled endpoint is generated from this value.

Subtract the rate from one to find what remains each period; do not subtract a fixed amount.

The retention factor is \(0.88\). Therefore \(A=5000(0.88)^{3}=3407.36\), matching the exact plotted endpoint.

Subtract the rate from one to find what remains each period; do not subtract a fixed amount.

The retention factor is \(0.75\). Therefore \(A=240(0.75)^{4}=75.94\), matching the exact plotted endpoint.

Subtract the rate from one to find what remains each period; do not subtract a fixed amount.

The retention factor is \(0.92\). Therefore \(A=18000(0.92)^{2}=15235.2\), matching the exact plotted endpoint.

Subtract the rate from one to find what remains each period; do not subtract a fixed amount.

The retention factor is \(0.5\). Therefore \(A=96(0.5)^{3}=12\), matching the exact plotted endpoint.

Subtract the rate from one to find what remains each period; do not subtract a fixed amount.

The retention factor is \(0.97\). Therefore \(A=1400(0.97)^{5}=1202.23\), matching the exact plotted endpoint.

Divide elapsed time by doubling time before using that quotient as the exponent.

The number of doubling intervals is \(t/d=9/3=3\). Thus \(A=25\cdot2^{3}=200\), matching the plotted values.

Divide elapsed time by doubling time before using that quotient as the exponent.

The number of doubling intervals is \(t/d=10/5=2\). Thus \(A=80\cdot2^{2}=320\), matching the plotted values.

Divide elapsed time by doubling time before using that quotient as the exponent.

The number of doubling intervals is \(t/d=12/4=3\). Thus \(A=12\cdot2^{3}=96\), matching the plotted values.

Divide elapsed time by doubling time before using that quotient as the exponent.

The number of doubling intervals is \(t/d=18/6=3\). Thus \(A=150\cdot2^{3}=1200\), matching the plotted values.

Divide elapsed time by doubling time before using that quotient as the exponent.

The number of doubling intervals is \(t/d=8/2=4\). Thus \(A=7\cdot2^{4}=112\), matching the plotted values.

Count half-life intervals with t divided by d, then halve multiplicatively rather than subtracting a fixed amount.

There are \(3\) half-life intervals. Therefore \(A=640(\frac12)^{3}=80\). Each plotted interval halves the current amount.

Count half-life intervals with t divided by d, then halve multiplicatively rather than subtracting a fixed amount.

There are \(3\) half-life intervals. Therefore \(A=200(\frac12)^{3}=25\). Each plotted interval halves the current amount.

Count half-life intervals with t divided by d, then halve multiplicatively rather than subtracting a fixed amount.

There are \(4\) half-life intervals. Therefore \(A=96(\frac12)^{4}=6\). Each plotted interval halves the current amount.

Count half-life intervals with t divided by d, then halve multiplicatively rather than subtracting a fixed amount.

There are \(3\) half-life intervals. Therefore \(A=1500(\frac12)^{3}=187.5\). Each plotted interval halves the current amount.

Count half-life intervals with t divided by d, then halve multiplicatively rather than subtracting a fixed amount.

There are \(4\) half-life intervals. Therefore \(A=448(\frac12)^{4}=28\). Each plotted interval halves the current amount.

Do not round t divided by d to a whole number; use the full quotient as the exponent.

The exponent is the possibly fractional interval count \(t/d=12/8=1.5\). Thus \(A=100(2)^{1.5}=282.84\).

Do not round t divided by d to a whole number; use the full quotient as the exponent.

The exponent is the possibly fractional interval count \(t/d=9/6=1.5\). Thus \(A=320(0.5)^{1.5}=113.14\).

Do not round t divided by d to a whole number; use the full quotient as the exponent.

The exponent is the possibly fractional interval count \(t/d=25/10=2.5\). Thus \(A=45(2)^{2.5}=254.56\).

Do not round t divided by d to a whole number; use the full quotient as the exponent.

The exponent is the possibly fractional interval count \(t/d=10/4=2.5\). Thus \(A=900(0.5)^{2.5}=159.1\).

Do not round t divided by d to a whole number; use the full quotient as the exponent.

The exponent is the possibly fractional interval count \(t/d=18/12=1.5\). Thus \(A=60(2)^{1.5}=169.71\).

Evaluate both complete models with full precision at the same time, then subtract and round once.

Model A gives \(2680.19\), and Model B gives \(2776.35\). Subtraction gives \(96.16\), so Model B is larger.

Evaluate both complete models with full precision at the same time, then subtract and round once.

Model A gives \(16791.47\), and Model B gives \(17193.43\). Subtraction gives \(401.96\), so Model B is larger.

Evaluate both complete models with full precision at the same time, then subtract and round once.

Model A gives \(881.17\), and Model B gives \(955.06\). Subtraction gives \(73.89\), so Model B is larger.

Evaluate both complete models with full precision at the same time, then subtract and round once.

Model A gives \(5970.78\), and Model B gives \(5878.08\). Subtraction gives \(92.71\), so Model A is larger.

Evaluate both complete models with full precision at the same time, then subtract and round once.

Model A gives \(358.32\), and Model B gives \(354.31\). Subtraction gives \(4.01\), so Model A is larger.

Isolate the power first, then use a source-aligned square/cube relationship or divide by the known factor power.

Divide to get (1+r)^2=1.21, so 1+r=1.1 and r=0.1=10%. Substitution in the original relationship verifies the result without requiring logarithms.

Isolate the power first, then use a source-aligned square/cube relationship or divide by the known factor power.

Divide to get (1-r)^2=0.81, so 1-r=0.9 and r=0.1=10%. Substitution in the original relationship verifies the result without requiring logarithms.

Isolate the power first, then use a source-aligned square/cube relationship or divide by the known factor power.

P=1458/(0.9)^3=2000. Substitution in the original relationship verifies the result without requiring logarithms.

Isolate the power first, then use a source-aligned square/cube relationship or divide by the known factor power.

P=2662/(1.1)^2=2200. Substitution in the original relationship verifies the result without requiring logarithms.

Isolate the power first, then use a source-aligned square/cube relationship or divide by the known factor power.

(1+r)^2=1.44, so 1+r=1.2 and r=20%. Substitution in the original relationship verifies the result without requiring logarithms.

Validate any proposed model at time zero and after one stated period or interval.

Use P(1.06)^t because each period keeps the original 100% and adds 6%. A valid model also returns the initial value at \(t=0\) and meets the stated one-interval behavior.

Validate any proposed model at time zero and after one stated period or interval.

Use P(0.88)^t because 88% remains after each 12% decrease. A valid model also returns the initial value at \(t=0\) and meets the stated one-interval behavior.

Validate any proposed model at time zero and after one stated period or interval.

Use t/d because the exponent counts how many doubling intervals have elapsed. A valid model also returns the initial value at \(t=0\) and meets the stated one-interval behavior.

Validate any proposed model at time zero and after one stated period or interval.

Each interval halves the amount then remaining, so the change is multiplicative. A valid model also returns the initial value at \(t=0\) and meets the stated one-interval behavior.

Validate any proposed model at time zero and after one stated period or interval.

Retain full precision for both values, subtract, and round only the requested final difference. A valid model also returns the initial value at \(t=0\) and meets the stated one-interval behavior.