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No; sometimes it will make it worse.

Heteroskedasticity where the spread is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity will often be made worse by that transformation.

Because taking logs "pulls in" more extreme values on the right (high values), while values at the far left (low values) tend to get stretched back:

this means spreads will become smaller if the values are large but may become stretched if the values are already small.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)

No; sometimes it will make it worse.

Heteroskedasticity where the spread* is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity will often be made worse by that transformation.

Because taking logs "pulls in" more extreme values on the right (high values), while values at the far left (low values) tend to get stretched back:

this means spreads will become smaller if the values are large but may become stretched if the values are already small.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)

* Here spread is some measure in the units of the data, like standard deviation, interquartile range, mean deviation, etc.

No; sometimes it will make it worse.

Heteroskedasticity where the spread is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity will often be made worse by that transformation.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)

No; sometimes it will make it worse.

Heteroskedasticity where the spread is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity will often be made worse by that transformation.

Because taking logs "pulls in" more extreme values on the right (high values), while values at the far left (low values) tend to get stretched back:

this means spreads will become smaller if the values are large but may become stretched if the values are already small.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)

No; sometimes it will make it worse.

Heteroskedasticity where the spread is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity may often be worse.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)

No; sometimes it will make it worse.

Heteroskedasticity where the spread is close to proportional to the conditional mean will tend to be improved by taking log(y), but if it's not increasing with the mean at close to that rate (or more), then the heteroskedasticity will often be made worse by that transformation.

If you know the approximate form of the heteroskedasticity, then you can sometimes work out a transformation that will approximately make the variance constant. This is known as a variance-stabilizing transformation; it is a standard topic in mathematical statistics. There are a number of posts on our site that relate to variance-stabilizing transformations.

If the spread is proportional to the square root of the mean (variance proportional to the mean), then a square root transformation - the variance-stabilizing transformation for that case - will tend to do much better than a log transformation; the log transformation does "too much" in that case. In the second plot we have the spread decrease as the mean increased, and then taking either logs or square roots would make it worse. (It turns out that the 1.5 power actually does reasonably well at stabilizing variance in that case.)